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NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

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NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability

Edited By Ravindra Pindel | Updated on Apr 26, 2023 03:00 PM IST | #CBSE Class 10th

NCERT exemplar Class 10 Maths solutions chapter 13 is provided to students to prepare for CBSE exam. It is an extension to the learnings of statistics and probability done in Class 9. The chapter on Statistics and Probability is of great importance for the examinations as well as future career prospects for a student. All the data analytics and artificial intelligence-related verticals require a sound knowledge of Statistics and Probability. The NCERT exemplar Class 10 Maths chapter 13 solutions are curated by our highly experienced content development team which enables the students to study and practice NCERT Class 10 Maths effectively.

These NCERT exemplar Class 10 Maths chapter 13 solutions follow the CBSE 10 Maths Syllabus Also, in this Chapter 13, there are exercises that revolve around various statistical measures, including mean, mode, and median. Through these exercises, students will gain an understanding of how to solve these types of problems. Additionally, the chapter delves into the topic of cumulative frequency distribution and cumulative frequency curves, providing further explanation and insight.

NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.1

Question:1

In the formula \bar{x}= a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}
for finding the mean of grouped data di’s are deviations from

(A) lower limits of the classes
(B) upper limits of the classes
(C) mid points of the classes
(D) frequencies of the class marks

Answer:

Answer. [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide
by the total number of observations.
We know that di = xi – a
where xi is data and a is mean
So, di are the derivative from mid-point of the classes.

Question:2

While computing mean of grouped data, we assume that the frequencies are
(A) evenly distributed over all the classes
(B) centred at the class marks of the classes
(C) centred at the upper limits of the classes
(D) centred at the lower limits of the classes

Answer:

Answer. [B]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
Hence while computing mean of grouped data, we assume that the frequencies are centered at the class marks of the classes

Question:3

If xi’s are the mid points of the class intervals of grouped data, fi’s are the corresponding frequencies and \bar{x} is the mean, then \Sigma\left(f_{i} x_{i}-\bar{x}\right) is equal to
(A) 0 (B) –1 (C) 1 (D) 2

Answer:

Answer. [A]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
That is mean \left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{n}
By cross multiplication we get
\sum f_{i}x_{i}= \bar{x}n\cdots (1)
\sum \left ( f_{i}x_{i} -\bar{x}\right )= \sum f_{i}x_{i} -\sum \bar{x}
= n\bar{x}-\sum \bar{x} (from equation (1))
= n\bar{x}-n\bar{x}
= 0

Question:4

In the formula \bar{x}= a+h\left ( \frac{\sum f_{i}u_{i}}{\sum f_{i}} \right ) for finding the mean of grouped frequency distribution, ui =
(A) \frac{x_{i}+a}{h} (B) h(xi – a) (C) \frac{x_{i}-a}{h} (D) \frac{a-x_{i}}{h}

Answer:

Answer. [C]
Solution. Mean: It is the average of the given numbers/observations. It is easy to calculate mean. First of all, add up all the observations and then divide by the total number of observations.
Also we know that di = xi – a and u_{i}= \frac{d_{i}}{h}
u_{i}= \frac{d_{i}}{h}
put di = xi – a
u_{i}= \frac{x_{i}-a}{h}
Hence option C is correct

Question:5

The abscissa of the point of intersection of the less than type and of the more than type cumulative frequency curves of a grouped data gives its
(A) mean (B) median (C) mode (D) all the three above

Answer:

Answer. [B]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
If we make graph of less than type and of more than type grouped data and find the intersection point then the value at abscissa is the median of the grouped data.
Hence option (B) is correct.

Question:6

For the following distribution :

Class

0-5

5-10

10-15

15-20

20-25

Frequency

10

15

12

20

9

the sum of lower limits of the median class and modal class is
(A) 15 (B) 25 (C) 30 (D) 35

Answer:

Answer. [B]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Class

Frequency

Cumulative Frequency (C.F)

0-5

10

10

5-10

15

(10+15=25)

10-15

12

(25+12=37)

15-20

20

(37+20=57)

20-25

9

(57+9=66)



N=66



\frac{N}{2}=\frac{66}{2}=33 which ies in the class 10-15.
Hence the median class is 10 – 15
The class with maximum frequency is modal class which is 15 – 20
The lower limit of median class = 10
The lower limit of modal class = 15
Sum = 10 + 15 = 25

Question:7

Consider the following frequency distribution:

Class

0-5

6-11

12-17

18-23

24-29

Frequency

13

10

15

8

11

The upper limit of the median class is
(A) 17 (B) 17.5 (C) 18 (D) 18.5

Answer:

Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.
Class in not continuous. So we have to make 1t continuous first.

Class

Frequency

Cumulative Frequency

0-5.5

13

13

5.5-11.5

10

(13+10=23)

11.5-17.5

15

(23+15=38)

17.5-23.5

8

(38+8=46)

23.5-29.5

11

(46+11=57)



N = 57



\frac{N}{2}= \frac{57}{2}= 28\cdot 5
Here the median class is 15.5 – 17.5
upper limit of median class is 17.5

Question:8

For the following distribution :

Marks

Number of students

Below 10
Below 20
Below 30
Below 40
Below 50
Below 60

3
12
27
57
75
80

the modal class is
(A) 10-20 (B) 20-30 (C) 30-40 (D) 50-60

Answer:

Answer. [C]
Solution.
Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to
summarize categorical variables.

Marks

Frequency

Cumulative Frequency

0-10

3

3

10-20

12-3=9

12

20-30

27-12=15

27

30-40

57-27=30

57

40-50

75-57=18

75

50-60

80-75=5

80



=80



The class with highest frequency is 30-40
Hence 30 – 40 is the modal class.

Question:9

Consider the data :

Class

65-85

85-105

105-125

125-145

145-165

165-185

185-205

Frequency

4

5

13

20

14

7

4

The difference of the upper limit of the median class and the lower limit of the modal class is
(A) 0 (B) 19 (C) 20 (D) 38

Answer:

Answer. [C]

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.


Class

Frequency

Cumulative Frequency

65-85

4

4

85-105

5

(4+5=9)

105-125

13

(9+13=22)

125-145

20

(22+20=42)

145-165

14

(42+14=56)

165-185

7

(56+7=63)

185-205

4

(63+4=67)



N = 67



\frac{N}{2}= \frac{67}{2}= 33\cdot 5
Median class = 125 – 145
upper limit of median = 145
The class with maximum frequency is modal class which is 125 – 145
lower limit of modal class = 125
Difference of the upper limit of median and lower limit of modal = 145 – 125 = 20

Question:10

The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below :

Class

13.8-14

14-14.2

14.2-14.4

14.4-14.6

14.6-14.8

14.8-15

Frequency

2

4

5

71

48

20

The number of athletes who completed the race in less than 14.6 seconds is :
(A) 11 (B) 71 (C) 82 (D) 130

Answer:

Answer. [C]
Solution. Frequency:- The number of times a event occurs is a specific period is called frequency.
The number of athletes who are below 14.6 = frequency of class (13.8-14) + frequency of class (14- 14.2) +
frequency of class (14.2-14.4) + frequency of class (14.4-14.6)
= 2 + 4 + 5 + 71 = 82
Hence the frequency of race completed in less than 14.6 = 82

Question:11

Consider the following distribution :

Marks obtained

Number of students

More than or equal to 0
More than or equal to 10
More than or equal to 20
More than or equal to 30
More than or equal to 40
More than or equal to 50


63
58
55
51
48
42

the frequency of the class 30-40 is
(A) 3 (B) 4 (C) 48 (D) 51

Answer:

Answer. [A]
Solution. Frequency distribution:
It tells how frequencies are distributed over values in a frequency distribution. However mostly we use
frequency distribution to summarize categorical variables.

Marks obtained

Cumulative Frequency

Frequency

0-10

63

5

10-20

58

3

20-30

55

4

30-40

51

3

40-50

48

6

50-60

42

42

So the frequency of class 30 – 40 is 3.

Question:12

If an event cannot occur, then its probability is
(A) 1 (B)\frac{3}{4} (C) \frac{1}{2} (D) 0

Answer:

Answer. [D]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Here number of favorable cases is 0.
Probability = \frac{Number \, of\, favourable\, case}{Total\, number\, of\, cases}
Probability = \frac{0}{\left ( Total\, cases \right )}= 0

Question:13

Which of the following cannot be the probability of an event?
(A)\frac{1}{3} (B) 0.1 (C) 3% (D)\frac{17}{16}

Answer:

Answer. [D]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
(A) 1/3
Here 0 < 1/3 < 1
Hence it can be the probability of an event.
(B) 0.1
Here 0 < 0.1 < 1
Hence it can be the probability of an event.
(C) 3% = 3/100 = 0.03
Here 0 < 0.03 < 1
Hence it can be the probability of an event.
(D)17/16
Here \frac{17}{16}> 1
Hence \frac{17}{16} is not a probability of event
Hence option (D) is correct answer.

Question:14

An event is very unlikely to happen. Its probability is closest to
(A) 0.0001 (B) 0.001 (C) 0.01 (D) 0.1

Answer:

Answer. [A]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The descending order of option (A), (B), (C), (D) is
0.1 > 0.01 > 0.001 > 0.0001 that is (D) > (C) > (B) > (A)
We can also say that it is the order of happening of an event.
Here 0.0001 it is the smallest one.
Hence 0.0001 is very unlikely to happen

Question:15

If the probability of an event is p, the probability of its complementary event will be
(A) p – 1 (B) p (C) 1 – p (D)1-\frac{1}{p}

Answer:

Answer. [C]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of an event = p
Let the probability of its complementary event = q
We know that total probability is equal to 1.
Hence, p + q = 1
q = 1 – p

Question:16

The probability expressed as a percentage of a particular occurrence can never be
(A) less than 100
(B) less than 0
(C) greater than 1
(D) anything but a whole number

Answer:

Answer. [B]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability expressed as presentage of an event A is btewwen 0 to 100.
Hence we can say that probability can never be less than 0.
Hence option (B) is correct.

Question:17

If P(A) denotes the probability of an event A, then
(A) P(A) < 0 (B) P(A) > 1 (C) 0 ≤ P(A) ≤ 1 (D) –1 ≤ P(A) ≤ 1

Answer:

Answer. [C]
Solution. Probability; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
(A) P(A) < 0
It is not represent the probability of event A because probability of an event can never be less than 0.
(B) P(A) > 1
It is not represent the probability of event A because probability of an event can never be greater than 1.
(C) 0 ≤ P(A) ≤ 1
It represent probability of event A because probability of an event is always lies from 0 to 1.
(D) –1 ≤ P(A) ≤ 1
It is not represent the probability of event A because probability of an event can never be equal to -1.
Hence option (C) is correct .

Question:18

A card is selected from a deck of 52 cards. The probability of its being a red face card is
(A) \frac{3}{26} (B) \frac{3}{13} (C) \frac{2}{13} (D) \frac{1}{2}

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 52
Red face cards = 6
Favorable cases = 6
Let event A is to select a card from 52 card.
Probability that it is a red card is p(A)
P\left ( A \right )= \frac{Number\, of\, favourable\, cases}{Total\, number\, of\, cases}
P\left ( A \right )=\frac{6}{52}= \frac{3}{26}

Question:19

The probability that a non leap year selected at random will contain 53 Sundays is
(A) \frac{1}{7} (B) \frac{2}{7} (C) \frac{3}{7} (D) \frac{5}{7}

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
In 365 days there are 52 weeks and 1 day.
If it contain 53 sunday then the 1 day of the year must be sunday.
But there are total 7 days.
Hence total number of favorable cases = 1
Hence probability of 53 sunday = \frac{Number\, of\, favourable\, cases}{Total\, cases}
= \frac{1}{7}

Question:20

When a die is thrown, the probability of getting an odd number less than 3 is
(A) \frac{1}{6} (B) \frac{1}{3} (C) \frac{1}{2} (D) 0

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total no. of cases = 6
odd number less than 3 = 1
Number of favorable cases = 1
Probability = = \frac{1}{6}

Question:21

A card is drawn from a deck of 52 cards. The event E is that card is not an ace of hearts. The number of outcomes favourable to E is
(A) 4 (B) 13 (C) 48 (D) 51

Answer:

Answer. [D]
Solution. Total number of cards = 52
Ace of hearts = 1
The card is not an ace of hearts = 52 – 1 = 51
The number of outcomes favourabe to E = 51

Question:22

The probability of getting a bad egg in a lot of 400 is 0.035. The number of bad eggs in the lot is
(A) 7 (B) 14 (C) 21 (D) 28

Answer:

Answer. [B]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Let event A is to get a bad egg.
So, p (A) = 0.035 (given)
P(A) = \frac{Number\, of\, favorable\ cases }{Total\, number\, of\, cases}
0.035 = \frac{Number\, of\, favorable\ cases }{400}
Number of favourable cases = \frac{35}{1000}\times 400= \frac{140}{10}= 14

Question:23

A girl calculates that the probability of her winning the first prize in a lottery is 0.08. If 6000 tickets are sold, how many tickets has she bought?
(A) 40 (B) 240 (C) 480 (D) 750

Answer:

Answer. [C]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases = 6000
Probability of getting first prize (p(A)) = 0.08
p(A) = \frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}
0.08 × 6000 = Number of tickets the bought
\frac{8}{100}\times 6000 = Number of tickets the bought
Number of tickets the bought = 480.

Question:24

One ticket is drawn at random from a bag containing tickets numbered 1 to 40. The probability that the selected ticket has a number which is a multiple of 5 is
(A) \frac{1}{5} (B) \frac{3}{5} (C) \frac{4}{5} (D) \frac{1}{3}

Answer:

Answer. [A]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total tickets = 40
Number of tickets multiple of 5 = 5, 10, 15, 20, 25, 30, 35, 40
Total favourable cases = 8
Let A be the event of getting a ticket with number multiple of 5.
p(A) = \frac{Number\, of\, tickets\, the\, bought}{Total\, tickets}
p\left ( A \right )= \frac{8}{40}= \frac{1}{5}

Question:25

Someone is asked to take a number from 1 to 100. The probability that it is a prime is
(A) \frac{1}{5} (B) \frac{6}{25} (C) \frac{1}{4} (D) \frac{13}{50}

Answer:

Answer. [C]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total number of cases = 100
prime number from 1 to 100 = 2, 3, 5, 7, 9, 11, 13, 17, 19,
23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 97
Total prime numbers from 1 to 100 = 25
Probability of getting prime number = \frac{prime\, no.\, from\, 1\, to\, 100}{Total\, number}
\\=\frac{25}{100}\\\\=\frac{1}{4}

Question:26

A school has five houses A, B, C, D and E. A class has 23 students, 4 from house A, 8 from house B, 5 from house C, 2 from house D and rest from house E. Asingle student is selected at random to be the class monitor. The probability that theselected student is not from A, B and C is
(A) \frac{4}{23} (B) \frac{6}{23} (C) \frac{8}{23} (D) \frac{17}{23}

Answer:

Answer. [B]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a
random event. The value is expressed from zero to one.
Total students = 23
Students in A, B, C = 4 + 8 + 5 = 17
Students in C, D = 23 – 17 = 6
Number of favourable cases = 6
Let A be the event that the student is not from A, B, C
p\left ( A \right )= \frac{Student\, from\, C,D}{Total\, students}
p\left ( A \right )= \frac{6}{23}

NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.2

Question:1

The median of an ungrouped data and the median calculated when the same data is grouped are always the same. Do you think that this is a correct statement? Give reason.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Ungrouped data:- The data which is not grouped is called ungrouped data.
The median is the middle number in the grouped data but when data is ungrouped the median is also changed.
Hence the median is not same of grouped and ungrouped data

Question:2

In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula
\bar{x}= a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}where a is the assumed mean. a must be one of the mid-points of the classes. the last statement correct? Justify your answer.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
Mean : It is the average of the given numbers. It is easy
to calculate mean. First of all add up all the numbers then divide by how many numbers are there.
This last statement is not correct because a can be any point in the grouped data it is not necessary that a must be mid-point.
Hence the statement is false.

Question:3

Is it true to say that the mean, mode and median of grouped data will always be different? Justify your answer.

Answer:

Answer. [False]
Solution. Mean : It is the average of the given numbers. It is easy to calculate mean. First of all add up all the numbers then divide by how many numbers are there.
Grouped data are data formed by aggregating individual observations of a variable into groups.
The mean, mode and median of grouped data can be the same it will depend on what type of data is given.
Hence the statement is false.

Question:4

Will the median class and modal class of grouped data always be different? Justify your answer.

Answer:

Answer. [False]
Solution. Grouped data are data formed by aggregating individual observations of a variable into groups.
The median is always the middle number and the modal class is the class with highest frequency it can be happen that the median class is of highest frequency.
So the given statement is false median class and mode class can be same.

Question:5

In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is \frac{1}{4} . Is this correct? Justify your answer.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total children = 3
Cases – GGG, GGB, GBG, BGG, BBB, BBG, BGB, GBB were G is girl and B is boy.
Probability = \frac{Number\, of\, favorable\ cases, }{Total\, number\, of\, cases}
Probability of 0 girl = \frac{1}{8}
Probability of 1 girl = \frac{3}{8}
Probability of 2 girl = \frac{3}{8}
Probability of 3 girl = \frac{1}{8}
Here they are not equal to \frac{1}{4}

Question:6

A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig.). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here 3 contain 50% of the region and 1, 2, contain 25%, 15% of the region.
Probability= \frac{Number\, of\, favorable\ cases }{Total\, number\, of\, cases}
Probability\, of\, 3= \frac{50}{100}= \frac{1}{2}
Probability\, of\, 1= \frac{25}{100}= \frac{1}{4}
probability\, of\, 2= \frac{25}{100}= \frac{1}{4}
All probabilities are not equal. So the given statement is false.

Question:7

Apoorv throws two dice once and computes the product of the numbers appearing on the dice. Peehu throws one die and squares the number that appears on it. Who has the better chance of getting the number 36? Why

Answer:

Answer. [Peehu]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
As apoorv throws two dice total cases = 36
Product is 36 when he get = (6, 6)
Number of favourable cases = 1
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that Apoorv get 36 = \frac{1}{36}
Peehu throws are die total cases = 6
Square of 6 is 36
Hence case = 1
Probability that Peehu get 36 = \frac{1}{6}
Hence Peehu has better cases to get 36.

Question:8

When we toss a coin, there are two possible outcomes - Head or Tail. Therefore, the probability of each outcome is \frac{1}{2} . Justify your answer.

Answer:

Answer. [True]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases when we toss a coin = 2(H, T)
Probability = \frac{Number\, of\, favourable\ cases, }{Total\, number\, of\, cases}
Probability of head = \frac{1}{2}
Probability of tail = \frac{1}{2}
Hence the probability of each outcome is \frac{1}{2} .


Question:9

A student says that if you throw a die, it will show up 1 or not 1. Therefore, the probability of getting 1 and
the probability of getting ‘not 1’ each is equal to \frac{1}{2} . Isthis correct? Give reasons.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Here total cases = 6
Number of favourable cases in getting 1 = 1
Probability = \frac{Number\, of\, favourable\ cases, }{Total\, number\, of\, cases}
Probability of getting 1= \frac{1}{6}
Number of favourable cases 'not 1' = 5 (2, 3, 4, 5, 6)
Probability of not 1 = \frac{5}{6}
Hence they are not equal to \frac{1}{2}

Question:10

I toss three coins together. The possible outcomes are no heads, 1 head, 2 heads and 3 heads. So, I say that probability of no heads is \frac{1}{4} . What is wrong with thisconclusion?

Answer:

Answer. [\frac{1}{8} ]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total cases in tossing three coins = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Number of case with no head = TTT
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of no head =\frac{1}{8}
The conclusion that probability of no head is \frac{1}{4} is wrong because as we calculate it above, it comes out \frac{1}{8} . Hence the probability of no head is \frac{1}{8}

Question:11

If you toss a coin 6 times and it comes down heads on each occasion. Can you say that the probability of getting a head is 1? Give reasons.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
The probability of getting a head is 1, means that we never get tail. But this is not true because we have both head and tail in a coin. Hence probability of getting head is 1 is false.

Question:12

Sushma tosses a coin 3 times and gets tail each time. Do you think that the outcome of next toss will be a tail? Give reasons.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because when we toss a coin we can get either tail or head and the probability of each is \frac{1}{2}.
So, it is not necessary that she gets tail at fourth toss. She can get head also.

Question:13

If I toss a coin 3 times and get head each time, should I expect a tail to have a higher chance in the 4th toss? Give reason in support of your answer.

Answer:

Answer. [False]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
No, because we get head or tail after tossing a coin that is the probability of both outcomes is \frac{1}{2} .
Hence tail is not have higher chance than head.
Both are have equal chance.

Question:14

A bag contains slips numbered from 1 to 100. If Fatima chooses a slip at random from the bag, it will either be an odd number or an even number. Since this situation has only two possible outcomes, so, the probability of each is \frac{1}{2} . Justify.

Answer:

Answer. [True]
Solution. Probability: probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Total slips = 100
Slips with even number = 50
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of slip with even number = \frac{50}{100}= \frac{1}{2}
Slips with odd number = 50
Probability of slip with odd number = \frac{50}{100}= \frac{1}{2}
Hence the probability of each is \frac{1}{2} .

NCERT Exemplar Class 10 Maths Solutions Chapter 13 Statistics and Probability: Exercise-13.3

Question:1

Find the mean of the distribution :

Class

1-3

3-5

5-7

7-10

Frequency

9

22

27

17

Answer:

Answer. [5.5]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

    Class

    Marks (xi)

    Frequency(fi)

    fixi

    1-3

    2

    9

    18

    3-5

    4

    22

    88

    5-7

    6

    27

    162

    7-10

    8.5

    17

    144.5





    \sum f_{i}= 75

    \sum f_{i}x_{i}= 412\cdot 5


    mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{412\cdot 5}{75}= 5\cdot 5

Question:2

Calculate the mean of the scores of 20 students in a mathematics test :

Marks

10-20

20-30

30-40

40-50

50-60

Number of students

2

4

7

6

1

Answer:

Answer. [35]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)
    4 Now divide sum of (fx) by sum of (f)

Marks

xi

No.of students fi

fixi

10-20

15

2

30

20-30

25

4

100

30-40

35

7

245

40-50

45

6

270

50-60

55

1

55

\sum f_{i}= 20 \sum f_{i}x_{i}= 700
mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{700}{20}= 35

Question:3

Calculate the mean of the following data

Class

4-7

8-11

12-15

16-19

Frequency

5

4

9

10

Answer:

Answer. [12.93]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

Class

xi

fi

fi xi

4-7

5.5

5

275

8-11

9.5

4

38

12-15

13.5

9

121.5

16-19

17.5

10

175





\sum f_{i}= 28

\sum f_{i}x_{i}= 362

mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{362}{28}= 12\cdot 93

Question:4

The following table gives the number of pages written by Sarika for completing her own book for 30 days :

no.of pages written per day

16-18

19-21

22-24

25-27

28-30

no.of days

1

3

4

9

13

Find the mean number of pages written per day.

Answer:

Answer. [26]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

    No.of pages written per day

    no.of days(fi)

    (xi)

    fixi

    16-18

    1

    17

    17

    19-21

    3

    20

    60

    22-24

    4

    23

    92

    25-27

    9

    26

    234

    28-30

    13

    29

    377



    \sum f_{i}= 30



    \sum f_{i}x_{i}= 780


    mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{780}{30}= 26

Question:5

The daily income of a sample of 50 employees are tabulated as follows :

Income (in Rs)

1-200

201-400

401-600

601-800

Number of employees

14

14

14

7

Find the mean daily income of employees.

Answer:

Answer. [356.5]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

    Income (in Rs )

    xi

    No.of employees

    fixi

    1-200

    100.5

    14

    1407

    201-400

    300.5

    15

    4507.5

    401-600

    500.5

    14

    7007

    601-800

    700.5

    7

    4903.5





    \sum f_{i}= 50

    \sum f_{i}x_{i}= 17825

    mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{17825}{50}= 356.5

Question:6

An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table :

no.of seats

100-104

104-108

108-112

112-116

116-120

Frequency

15

20

32

18

15

Determine the mean number of seats occupied over the flights.

Answer:

Answer. [109]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

Number of seats

Frequency fi

xi

fixi

100-104

15

102

1530

104-108

20

106

2120

108-112

32

110

3520

112-116

18

114

2052

116-120

15

118

177065268



\sum f_{i}= 100



\sum f_{i}x_{i}= 10992

mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{10992}{100}= 109\cdot 92
number of seats = 109

Question:7

The weights (in kg) of 50 wrestlers are recorded in the following table :

Weight (in Kg)

100-110

110-120

120-130

130-140

140-150

Number of wrestlers

4

14

21

8

3

Find the mean weight of the wrestlers.

Answer:

Answer. [123.4]
Solution.
Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

    Weight

    fi

    xi

    fixi

    100-110

    4

    105

    420

    110-120

    14

    115

    1610

    120-130

    21

    125

    2625

    130-140

    8

    135

    1080

    140-150

    3

    145

    435



    \sum f_{i}= 50



    \sum f_{i}x_{i}=6170


    mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{6170}{50}= 123\cdot 4 \ \ kg

Question:8

The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below :

MIleage (km/I)

10-20

12-14

14-16

16-18

Number of cars

7

12

18

13

Find the mean mileage.
The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?

Answer:

Answer. [14.48]
Solution. Here we calculate mean by following
1.Find the mid point of each interval.
2.Multiply the frequency of each interval by its mid point.
3.Get the sum of all the frequencies (f) and sum of all the (fx)
4.Now divide sum of (fx) by sum of (f)

MIleage (km/I)

No.of cars (fi)

xi

fixi

10-12

7

11

77

12-14

12

13

156

14-16

18

15

270

16-18

13

17

221



\sum f_{i}= 50


\sum f_{i}x_{i}= 724

mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{724}{50}= 14\cdot 48 \ \ km/L

Question:9

The following is the distribution of weights (in kg) of 40 persons :

Weight (in kg)

40-45

45-50

50-55

55-60

60-65

65-70

70-75

70-75

Number of person

4

4

13

5

6

5

2

1

Construct a cumulative frequency distribution (of the less than type) table for the data above.

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

CI.

f

cf

40-45

4

4

45-50

4

8

50-55

13

21

55-60

5

26

60-65

6

32

65-70

5

37

70-75

2

39

75-80

1

40

Question:10

The following table shows the cumulative frequency distribution of marks of 800 students in an examination:

Marks

Number of students

Below 10
Below 20
Below 30
Below 40
Below 50
Below 60
Below 70
Below 80
Below 90
Below 100

10
50
130
270
440
570
670
740
780
800

Construct a frequency distribution table for the data above.

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a
frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Marks

cf

f

0-10

10

10

10-20

50

50-10=40

20-30

130

130-50=80

30-40

270

270-130=140

40-50

440

440-270=170

50-60

570

570-440=130

60-70

670

670-570=100

70-80

740

740-670=70

80-90

780

780-740=40

90-100

800

800-780=20

Question:11

Form the frequency distribution table from the following data :

Marks (out of 90)

Number of candidates

More than or equal to 80
More than or equal to 70
More than or equal to 60
More than or equal to 50
More than or equal to 40
More than or equal to 30
More than or equal to 20
More than or equal to 10
More than or equal to 0

4
6
11
17
23
27
30
32
34

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

class

f

0-10

34-32=2

10-20

32-30=2

20-30

30-27=3

30-40

27-23=4

40-50

23-17=6

50-60

17-11=6

60-70

11-6=5

70-80

6-4-=2

80-90

4

Question:12

Find the unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class :

Height (in cm)

Frequency

Cumulative frequency

150-155
155-160
160-165
165-170
170-175
175-180

12
b
10
d
e
2

a
25
c
43
48
f

Total

50



Answer:

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to
summarize categorical variables.
a= 12 ( because first term of frequency and cumulative frequency is same )
12 + b = 25
b = 25 – 12
b = 13

25 + 10 = c
35= c
c + d = 43
35 + d = 43
d = 43 – 35
d=8

43 + e = 48
e = 48 – 43
e =5
48+2 = f
50 = f
Ans. a = 12, b = 13, c = 35, d = 8, e = 5, f = 50

Question:13

The following are the ages of 300 patients getting medical treatment in a hospital on a particular day :

Age (in yeras)

10-20

20-30

30-40

40-50

50-60

60-70

Number of patients

60

42

55

70

53

20

(i) Less than type cumulative frequency distribution.
(ii) More than type cumulative frequency distribution

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Age (in year)

No.of patients

less than 10

0

less than 20

60+0 = 60

less than 30

42+60 = 102

less than 40

102+55 =157

less than 50

157+70 = 227

less than 60

227+53 =280

less than 70

280 +20 =300

Age (in year)

No.of patients

More than or equal to 10
More than or equal to 20
More than or equal to 30
More than or equal to 40
More than or equal to 50
More than or equal to 60

60+42+55+70+53+20 = 300
42+55+70+53+20 = 240
55+70+53+20= 198
70+53+20 = 143
53+20 = 73
20








Question:14

Given below is a cumulative frequency distribution showing the marks secured by 50 students of a class :

Marks

Below 20

Below 40

Below 60

Below 80

Below 100

Number of students

17

22

29

37

50

Form the frequency distribution table for the data

Answer:

Solution. Frequency distribution: It tells how frequencies are distributed overvalues in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Marks

Number of students CF

f

0- 20

17

17

20- 40

22

22-17 = 5

40- 60

29

29 -22 = 7

60- 80

37

37-29 = 8

80-100

50

50 -37 = 13

Question:15

Weekly income of 600 families is tabulated below :

Weekly income (in Rs)

Number of families

0-1000
1000-2000
2000-3000
3000-4000
4000-5000
5000-6000

250
190
100
40
15
5

Total

600

Compute the median income.

Answer:

Answer. [1263.15]
Solution. n = 600
\frac{n}{2}= \frac{600}{2}= 300
l\iota= 1000, l = 1000, cf = 250, f = 190
median = \iota +\left ( \frac{\frac{n}{2}-cf}{f} \right )\times h
= 1000+\frac{\left ( 300-250 \right )}{190}\times 1000
= 1000+\frac{50}{190}\times 1000
= 1000+\frac{5000}{19}
= \frac{19000+5000}{19}= \frac{24000}{19}= 1263\cdot 15
Median = 1263.15

Question:16

The maximum bowling speeds, in km per hour, of 33 players at a cricket coaching centre are given as follows :

Speed (km/h)

85-100

100-115

115-130

130-145

Number of players

11

9

8

5

Calculate the median bowling speed.

Answer:

Answer. [109.16]
Solution.
Here n = 33
\frac{n}{2}= \frac{33}{2}= 16\cdot 5
\iota = 100
h = 15, f = 9 , cf = 11
Median = \iota +\left ( \frac{\frac{n}{2}-cf}{f} \right )\times h
=100+\frac{\left ( 16\cdot 5-11 \right )\times 15}{9}
=100+\left ( \frac{55}{10\times 9} \right )\times 15
=100+\frac{55\times 5}{30}
=100+\frac{275}{30}
= 100 + 9.16 \Rightarrow 109.16

Question:17

The monthly income of 100 families are given as below :

Income (in Rs)

Number of families

0-5000
5000-10000
10000-15000
15000-20000
20000-25000
25000-30000
30000-35000
35000-40000

8
26
41
16
3
3
2
1


Calculate the modal income.

Answer:

Answer. [11875]
Solution. Here l = 10000, f1 = 41, f0 = 26, f2 = 16, h = 5000
Mode = \iota +\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h
=1000+\left ( \frac{41-26}{2\times 41-26-16} \right )\times 5000
=1000+\frac{15}{40}\times 5000
= 10000 + 1875 = 11875
Modal income is 11875 Rs.

Question:18

The weight of coffee in 70 packets are shown in the following table :

Weight (in g)

Number of packets

200-201
201-202
202-203
203-204
204-205
205-206

12
26
20
9
2
1

Determine the modal weight.

Answer:

Answer. [201.7 g]
Solution.
Here l = 201, f1 = 26, f0 = 12, f2 = 20, h = 1
\Rightarrow 201+\left ( \frac{26-12}{2\times 26-12-20} \right )\times 1
\Rightarrow 201+\left ( \frac{14}{52-32} \right )
\Rightarrow 201+\frac{14}{20}
\Rightarrow 201+0\cdot 7= 201\cdot 7 g

Question:19

Two dice are thrown at the same time.
(i) Find the probability of getting same number on both dice.
(ii) Find the probability of getting different numbers on both dice.

Answer:

Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases after thrown of two dice = 36
(i) Same number = (1, 1), (2, 2), (3, 3), (4. 4), (5, 5), (6, 6)
Same number cases = 6
Let A be the event of getting same number.
Probability [p(A)] = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
p\left ( A \right )= \frac{6}{36}= \frac{1}{6}
(ii) Different number cases = 36 – same number case
= 36 –6 = 30
Let A be the event of getting different number
Probability [p(A)]= \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
p\left ( A \right )= \frac{30}{36}= \frac{5}{6}

Question:20

Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is
(i) 7?
(ii) a prime number?
(iii) 1?

Answer:

(i) Answer. [1/6]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases after throwing of two dice = 36
Cases when total is 7 = (1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2)
Total cases = 6
Let A be the event of getting total 7
Probability [p(A)]= \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 7 = \frac{6}{36}= \frac{1}{6}

(ii) Answer. [5/12]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Prime number as a sum = (1, 1), (1, 2), (2, 1), (1, 4),
(4, 1), (2, 3), (3, 2), (1, 6), (6, 1), (3, 4), (4, 3), (2, 5), (5, 2), (6, 5), (5, 6)
Cases = 15
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that sum is a prime number = \frac{15}{36}= \frac{5}{12}
(iii) Answer. [0]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
pairs from which we get sum 1 = 0
Cases = 0
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 1 = \frac{0}{36}= 0

Question:21

Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is
(i) 6
(ii) 12
(iii) 7

Answer:

(i) Answer. [1/9]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
For getting product 6 = (1, 6,), (6, 1), (2, 3), (3, 2)
Cases = 4
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting product 6 =\frac{4}{36}= \frac{1}{9}

(ii) Answer. [1/9]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
product 12 = (2, 6), (6, 2), (3, 4), (4, 3)
Cases = 4
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting product 12 =\frac{4}{36}= \frac{1}{9}

(iii) Answer. [0]
Solution. Probability ; probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Product 7 = 0 (case)
Cases = 0
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting product 7 =\frac{0}{36}= 0

Question:22

Two dice are thrown at the same time and the product of numbers appearing on them is noted. Find the probability that the product is less than 9.

Answer:

Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases in throwing two dice = 36
Product less than 9 cases = (1, 1). (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (4, 1), (4, 2), (5, 1), (6, 1)
Number of favourable cases = 16
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting product less than 9 = \frac{16}{36}= \frac{4}{9}

Question:23

Two dice are numbered 1, 2, 3, 4, 5, 6 and 1, 1, 2, 2, 3, 3, respectively. They are thrown and the sum of the numbers on them is noted. Find the probability of getting each sum from 2 to 9 separately.

Answer:

Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number of cases = 36
case of getting sum 2 = ( 1 , 1 ) ( 1 , 1 )
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 2 = \frac{2}{36}= \frac{1}{18}
case of getting sum 3 = (1, 2), (1, 2), (2, 1), (2, 1)
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 3= \frac{4}{36}= \frac{1}{9}
case of getting sum 4 = (1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 4= \frac{6}{36}= \frac{1}{6}
case of getting sum 5 = (2, 3), (2, 3), (4, 1),(4,1) (3, 2), (3, 2)
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 5 = \frac{6}{36}= \frac{1}{6}
case of getting sum 6 = (3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)
probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 6= \frac{6}{36}= \frac{1}{6}
case of getting sum 7 = (4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)
probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 7= \frac{6}{36}= \frac{1}{6}
case of getting sum 8 = (5, 3), (5, 3), (6, 2), (6, 2)
probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 8= \frac{4}{36}= \frac{1}{9}
case of getting sum 9 = (6, 3), (6, 3)
probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting sum 9= \frac{2}{36}= \frac{1}{18}

Question:24

A coin is tossed two times. Find the probability of getting at most one head.

Answer:

Answer. [3/4]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 4 (HH, TT, HT, TH)
Cases of at most 1 head = HT, TH, TT
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting at most 1 head = \frac{3}{4}

Question:25

A coin is tossed 3 times. List the possible outcomes. Find the probability of getting
(i) all heads
(ii) at least 2 heads

Answer:

(i) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Possible outcomes = (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Total cases = 8
Cases of getting all heads = (HHH)
Number of favourable cases = 1
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting all heads = \frac{1}{8}

(ii) Answer.[1/2]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one.
Possible outcomes = 8 (HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Cases of getting at least 2 heads = (HHH, HHT, HTH, THH)
Favorable cases = 4
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting at least 2 heads = \frac{4}{8}= \frac{1}{2}

Question:26

Two dice are thrown at the same time. Determine the probability that the difference of the numbers on the two dice is 2.

Answer:

Answer. [2/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Cases of getting difference 2 = (1, 3), (3, 1), (2, 4), (4, 2), (3, 5), (5, 3), (4, 6), (6, 4)
Favourable cases = 8
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting difference 2 = \frac{8}{36}= \frac{2}{9}

Question:27

A bag contains 10 red, 5 blue and 7 green balls. A ball is drawn at random. Find the probability of this ball being a
(i) red ball
(ii) green ball
(iii) not a blue ball

Answer:

(i) Answer. [5/11]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Red balls = 10
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting red ball = \frac{10}{22}= \frac{5}{11}

(ii) Answer. [7/22]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Green balls = 7
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting green ball = \frac{7}{22}

(iii) Answer. [17/22]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = 10 + 5 + 7 = 22
Not a blue ball = 22 – (blue ball)
= 22 – 5 = 17
robability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting not a blue ball = \frac{17}{22}

Question:28

The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. Determine the probability that the card is
(i) a heart
(ii) a king

Answer:

(i) Answer. [13/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 ( three cards are removed)
= 49
Total hearts = 13
Favourable cases = 13
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting a heart = \frac{13}{49}

(ii) Answer. [3/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 ( three cards are removed)
= 49
Total king = 4 – 1 = 3 ( 1 king is removed)
favourable cases = 3
bability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting a King= \frac{3}{49}

Question:29

The king, queen and jack of clubs are removed from a deck of 52 playing cards and then well shuffled. Now one card is drawn at random from the remaining cards. What is the probability that the card is
(i) a club
(ii) 10 of hearts

Answer:

(i) Answer. [10/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 = 49 ( three cards are removed)
Total club = 13 – 3 = 10 ( 3 club cards are removed)
favourable cases = 10
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting a club = \frac{10}{49}

(ii) Answer. [1/49]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 3 = 49 ( three cards are removed)
10 of heart = 1
favourable cases = 1
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting a heart = \frac{1}{49}

Question:30

All the jacks, queens and kings are removed from a deck of 52 playing cards. The remaining cards are well shuffled and then one card is drawn at random. Giving ace a value 1 similar value for other cards, find the probability that the card has a value
(i) 7
(ii) greater than 7
(iii) less than 7

Answer:

(i) Answer. [1/10]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 ( 12 cards are removed)
card with number 7 = 4
favourable cases = 4
probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting card 7= \frac{4}{10}= \frac{1}{10}

(ii) Answer. [3/10]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 (\mathbb{Q} 12 cards are removed)
Cards greater than 7 =8,9,10 (3 × 4 = 12)
favourable cases = 12
probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting card 7= \frac{12}{40}= \frac{3}{10}
(iii) Answer. [3/5]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 52 – 12 = 40 (\because 12 cards are removed)
Cards less than 7 = 1, 2, 3, 4, 5, 6 (6 × 4 = 24)
favourable cases = 24
probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting card 7= \frac{24}{40}= \frac{6}{10}= \frac{3}{5}

Question:31

An integer is chosen between 0 and 100. What is the probability that it is
(i) divisible by 7 ?
(ii) not divisible by 7?

Answer:

(i) Answer. [14/99]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number = 99 (between 0 to 100)
Number divisible by 7 = (7, 14, 21, 28,35, 42, 49, 56, 63, 70, 77, 84, 91, 98)
Favourable cases = 14
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting number divisible by 7 = \frac{14}{99}

(ii) Answer. [85/99]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number between 0 to 100 = 99
Number divisible by 7 = (7, 14, 21, 28,35, 42, 49, 56, 63, 70, 77, 84, 91, 98)
Favourable cases = 14
robability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting number divisible by 7 = \frac{14}{99}
= 1-\frac{14}{99}
= \frac{99-14}{99}= \frac{85}{99}

Question:32

Cards with numbers 2 to 101 are placed in a box. A card is selected at random. Find the probability that the card has
(i) an even number
(ii) a square number

Answer:

(i) Answer. [1/2]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total numbers from 2 to 101 = 100
Total even numbers from 2 to 101 = 50
Favourable cases = 50
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that card is with even number = \frac{50}{100}= \frac{1}{2}

(ii) Answer. [9/100]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total number from 2 to 101 = 100
Square numbers from 2 to 101 = (4, 9, 16, 25, 36, 49, 64, 81, 100)
Favourable cases = 9
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that the card is with a square number = \frac{9}{100}

Question:33

A letter of English alphabets is chosen at random. Determine the probability that the letter is a consonant.

Answer:

Answer. [21/26]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total alphabets = 26
Total consonant = 21
Favourable cases = 21
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that alphabet is consonant = \frac{21}{26}

Question:34

There are 1000 sealed envelopes in a box, 10 of them contain a cash prize of Rs 100 each, 100 of them contain a cash prize of Rs 50 each and 200 of them contain a cash prize of Rs 10 each and rest do not contain any cash prize. If they are well shuffled and an envelope is picked up out, what is the probability that it contains no cash prize ?

Answer:

Answer. [0.69]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total envelopes = 1000
Envelopes with no cash prize = Total envelopes – envelopes with cash prize
= 1000 – 10 – 100 – 200 = 690
Favourable cases = 690
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that the envelope is no cash prize = \frac{690}{1000}= \frac{69}{100}= 0\cdot 69

Question:35

Box A contains 25 slips of which 19 are marked Rs 1 and other are marked Rs 5 each. Box B contains 50 slips of which 45 are marked Rs 1 each and others are marked Rs 13 each. Slips of both boxes are poured into a third box and reshuffled. A slip is drawn at random. What is the probability that it is marked other than Rs 1 ?

Answer:

Answer. [11/75]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total slips = 25 + 50 = 75
Slips marked other than 1 = Rs. 5 slips + Rs. 13 slips
= 6 + 5 = 11
Favourable cases = 11
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that slips is not marked 1 = \frac{11}{75}

Question:36

A carton of 24 bulbs contain 6 defective bulbs. One bulbs is drawn at random. What is the probability that the bulb is not defective? If the bulb selected is defective and it is not replaced and a second bulb is selected at random from the rest, what is the probability that the second bulb is defective?

Answer:

Answer. [5/23]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total bulbs = 24
Defective = 6
not defective = 18
Probability that the bulb is not defective = \frac{18}{24}= \frac{3}{4}
Let the bulbs is defective and it is removed from 24 bulb.
Now bulbs remain = 23
In 23 bulbs, non-defective bulbs = 18
defective = 5
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Now probability that the bulb is defective = \frac{5}{23} .

Question:37

A child’s game has 8 triangles of which 3 are blue and rest are red, and 10 squares of which 6 are blue and rest are red. One piece is lost at random. Find the probability that it is a
(i) triangle
(ii) square
(iii) square of blue colour
(iv) triangle of red colour

Answer:

(i) Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 8 + 10 = 18
Total triangles = 8
Favourable cases = 8
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that piece is a triangle = \frac{8}{18}= \frac{4}{9}

(ii) Answer. [5/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 8 + 10 = 18
Total square = 10
Favourable cases = 10
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that the piece is a square = \frac{10}{18}= \frac{5}{9}

(iii) Answer. [1/3]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 10 + 8 = 18
Square of blue color = 6
favourable cases = 6
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that piece is a square of blue color = \frac{6}{18}= \frac{1}{3}

(iv) Answer. [5/18]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total piece = 10 + 8 = 18
triangle of red color = 8 – 3 = 5
favourable cases = 5
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that piece is a triangle of red colour = \frac{5}{18}

Question:38

In a game, the entry fee is Rs 5. The game consists of a tossing a coin 3 times. If one or two heads show, Shweta gets her entry fee back. If she throws 3 heads, she receives double the entry fees. Otherwise she will lose. For tossing a coin three times, find the probability that she
(i) loses the entry fee.
(ii) gets double entry fee.
(iii) just gets her entry fee.

Answer:

(i) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which the lose entry = 8 – (in which she gets entry book + in which she gets double)
= 8 – 6 (HHT, HTH, THH, TTH, THT, HTT) – 1(HHH)
= 8 – 7 = 1
Favourable cases = 1
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that she will lose money = \frac{1}{8}

(ii) Answer. [1/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which she gets double entry = HHH
favourable cases = 1
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that she gets double entry fee = \frac{1}{8}

(iii) Answer. [3/4]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
case in which she gets entry book = 6(HHT, HTH, THH, TTH, THT, HTT)
favourable cases = 6
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that she gets entry fees = \frac{6}{8}= \frac{3}{4}

Question:39

A die has its six faces marked 0, 1, 1, 1, 6, 6. Two such dice are thrown together and the total score is recorded.
(i) How many different scores are possible?
(ii) What is the probability of getting a total of 7?

Answer:

(i) Answer. [6]
Solution. Count the number of sums we can notice by using two dice of (0, 1, 1, 1, 6, 6) type.
We can get a sum of 0 = (0,0)
We can get a sum of 1 = (0,1) , (1,0)
We can get a sum of 2 = (1,1)
We can get a sum of 6 = (0,6) , (6,0)
We can get a sum of 7 = (6,1) , (1,6)
We can get a sum of 12 = (6,6)
We can get a score of 0, 1, 2, 6, 7, 12
Hence we can get 6 different scores.

(ii) Answer. [4/9]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cases = 36
Case of getting sum 7 = (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (1, 6), (6,1), ( 6,1), (6,1), ( 6,1), (6,1), (6,1),
Number of favourable cases = 12
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting a total 7 = \frac{12}{36}= \frac{1}{3}

Question:40

A lot consists of 48 mobile phones of which 42 are good, 3 have only minor defects and 3 have major defects. Varnika will buy a phone if it is good but the trader will only buy a mobile if it has no major defect. One phone is selected at random from the lot. What is the probability that it is
(i) acceptable to Varnika?
(ii) acceptable to the trader?

Answer:

(i) Answer. [7/8]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total mobiles = 48
Minor defective = 3
major defective = 3
good = 42
Varnika buy only good so favourable cases = 42
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that acceptable to Varnika = \frac{42}{48}= \frac{7}{8}
(ii) Answer. [15/16]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
total mobiles = 48
good = 42
minor defect = 3
major defect = 3
trader accept only good and minor defect.
So favourable cases = 42 + 3 = 45
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that trader accept \frac{45}{48}= \frac{15}{16}

Question:41

A bag contains 24 balls of which x are red, 2x are white and 3x are blue. A ball is selected at random. What is the probability that it is
(i) not red?
(ii) white?

Answer:

(i) Answer. [5/6]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = red + white + blue
24 = x + 2x + 3x
6x = 24
x = 4
Red balls = x = 4
White balls = 2x = 2 × 4 = 8
Blue balls = 3x = 3 × 4 = 12
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that ball is not red = \frac{blue+white}{24}
= \frac{8+12}{24}= \frac{20}{24}= \frac{5}{6}

(ii) Answer. [1/3]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total balls = red + white + blue
24 = 6x
x = 4
white balls = 2x = 2 × 4 = 8
Favourable cases = 8
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability of getting on white ball = \frac{8}{24}= \frac{1}{3}

Question:42

At a fete, cards bearing numbers 1 to 1000, one number on one card, are put in a box. Each player selects one card at random and that card is not replaced. If the selected card has a perfect square greater than 500, the player wins a prize. What is the probability that
(i) the first player wins a prize?
(ii) the second player wins a prize, if the first has won?

Answer:

(i) Answer. [0.009]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Total cards = 1000
Player wins prize with cards = (529, 576, 625, 676, 729, 784, 841, 900, 961)
Favourable cases = 9
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that player wins = \frac{9}{1000}= 0\cdot 009

(ii) Answer. [0.008]
Solution. Probability; Probability means possibility. It is a branch of mathematics that deals with the occurrence of a random event. The value is expressed from zero to one
Now the total cards are = 1000 – 1 = 999
Now the total winning cards = 9 – 1 = 8
Probability = \frac{Number\, of\, favourable\ cases }{Total\, number\, of\, cases}
Probability that second player wins after first = \frac{8}{999}\approx 0\cdot 008

Question:1

Find the mean marks of students for the following distribution

Marks

Number of students

0 and above
10 and above
20 and above
30 and above
40 and above
50 and above
60 and above
70 and above
80 and above
90 and above
100 and above

80
77
72
65
55
43
28
16
10
8
0

Answer:

Answer. [51.75]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

    Marks

    xi

    cf

    fi

    fixi

    0-10

    5

    80

    80-77 = 3

    15

    10-20

    15

    77

    787-72 = 5

    75

    20-30

    25

    72

    72-65 =7

    175

    30-40

    35

    65

    65-55 = 10

    350

    40-50

    45

    55

    55-43 = 12

    540

    50-60

    55

    43

    43-28 = 15

    825

    60-70

    65

    28

    28-16 = 12

    780

    70-80

    75

    16

    16-10 =6

    450

    80-90

    85

    10

    10-8 = 2

    170

    90-100

    95

    8

    8-0 = 8

    760







    \sum f_{i}= 80

    \sum f_{i}x_{i}= 4140

    mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{4140}{80}= 51\cdot 75

Question:2

Determine the mean of the following distribution :

Marks

Number of students

Below 10
Below 20
Below 30
Below 40
Below 50
Below 60
Below 70
Below 80
Below 90
Below 100

5
9
17
29
45
60
70
78
83
85

Answer:

Answer. [48.4]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

    Marks

    xi

    cf

    fi

    fixi

    0-10

    5

    5

    5

    15

    10-20

    15

    9

    9-5 =4

    75

    20-30

    25

    17

    17-9 = 8

    175

    30-40

    35

    29

    29-17 = 12

    350

    40-50

    45

    45

    45-29 = 16

    540

    50-60

    55

    60

    60-45 = 15

    825

    60-70

    65

    70

    70-60 = 10

    780

    70-80

    75

    78

    78-70 = 8

    450

    80-90

    85

    83

    83-78 = 5

    170

    90-100

    95

    85

    85-83 = 2

    760







    \sum f_{i}= 85

    \sum f_{i}x_{i}= 4115


    mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{4115}{85}= 48\cdot 4

Question:3

Find the mean age of 100 residents of a town from the following data :

Age equal and above (in years)

0

10

20

30

40

50

60

70

Number of Persons

100

90

75

50

25

15

5

0

Answer:

Answer. [31]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

    Marks

    xi

    fi

    0-10

    5

    100-90 =10

    50

    10-20

    15

    90-75 = 15

    225

    20-30

    25

    75-50 = 25

    625

    30-40

    35

    50-25 =25

    875

    40-50

    45

    25-15 =10

    450

    50-60

    55

    15-5 = 10

    550

    60-70

    65

    5-0 = 5

    325





    \sum f_{i}= 100

    \sum f_{i}x_{i}= 3100

    mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{3100}{100}= 31

Question:4

The weights of tea in 70 packets are shown in the following table :

Weight (in gram)

Number of packets

200-201
201-202
202-203
203-204
204-205
205-206

13
27
18
10
1
1

Find the mean weight of packets.

Answer:

Answer. [201.95]
Solution. Here we calculate mean by following the given steps:

  1. Find the mid point of each interval.

  2. Multiply the frequency of each interval by its mid point.

  3. Get the sum of all the frequencies (f) and sum of all the (fx)

  4. Now divide sum of (fx) by sum of (f)

    Weight

    xi

    fi

    fixi

    200-201

    200.5

    13

    2606.5

    201-202

    201.5

    27

    5440.5

    202-203

    202.5

    18

    3645.0

    203-204

    203.5

    10

    2035.0

    204-205

    204.5

    1

    204.5

    205-206

    205.5

    1

    205.5





    \sum f_{i}= 70

    \sum f_{i}x_{i}= 14137


    mean\left ( \bar{x} \right )= \frac{\sum f_{i}x_{i}}{\sum f_{i}}= \frac{14137}{70}= 201\cdot 95

Question:5

The weights of tea in 70 packets are shown in the following table :

Weight (in gram)

Number of packets

200-201
201-202
202-203
203-204
204-205
205-206

13
27
18
10
1
1

Draw the less than type ogive for this data and use it to find the median weight.

Answer:

Answer. [201.8]
Solution.

Weight

cf

Less than 201

13

Less than 202

27+13=40

Less than 203

40+18=58

Less than 204

58+10=68

Less than 205

68+1 =69

Less than 206

69+1 = 70


n = 70,\frac{n}{2}= \frac{70}{2}= 35
Hence the median is 201.8

Question:6

The weights of tea in 70 packets are shown in the following table :

Weight (in gram)

Number of packets

200-201
201-202
202-203
203-204
204-205
205-206

13
27
18
10
1
1

Draw the less than type and more than type ogives for the data and use them to find the median weight.

Answer:

Answer. [201.8]
Solution.

Less than type

More than type



Weight

Number of packets

Number of packets

Number of students

Less than 200

0

More than or equal to 200

70

Less than 201

13

More than or equal to 201

70-13 = 57

Less than 202

40

More than or equal to 202

57-27 =30

Less than 203

58

More than or equal to 203

30-18 =12

Less than 204

68

More than or equal to 204

12-10 = 2

Less than 205

69

More than or equal to 205

2-1 = 1

Less than 206

70

More than or equal to 206

1-1 = 0


Here\, \, n = 70,\frac{n}{2}= \frac{70}{2}= 35
Hence median = 201.8

Question:7

The table below shows the salaries of 280 persons.

Salary(in thousand (Rs))

Number of persons

5-10
10-15
15-20
20-25
25-30
30-35
35-40
40-45
45-50

49
133
63
15
6
7
4
2
1

Calculate the median and mode of the data.

Answer:

Solution.

Salary

fi

cf

5-10

49

49

10-15

133

49+133=182

15-20

63

182+63=245

20-25

15

245+15 = 260

25-30

6

260+6 = 266

30-35

7

266+7 = 273

35-40

4

273+4 = 277

40-45

2

277+2 = 279

45-50

1

279+1 = 280

n= 280,\frac{n}{2}= \frac{280}{2}= 140
f1 = 49, fm= 133, f2= 63, cf = 49, f = 133
l = 10, h = 5
median = \iota +\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h
=10+\frac{\left ( 140-49 \right )}{133}\times 5
=10+\frac{91\times 5}{133}
=10+\frac{455}{133}= 10+3\cdot 421
= 13\cdot 421
In thousands = 13.421 × 1000 = 13421 Rs.
Mode = \iota +\left [ \frac{f_{m}-f_{i}}{2f_{m}-f_{i}-f_{2}} \right ]\times h

=10+\left [ \frac{133-49}{2\times 133-49-63} \right ]\times 5

=10+\frac{84\times 5}{266-112}=10+\frac{84\times 5}{154}
=10 + 2.727
=12.727
In thousands = 12.727 × 1000 = 12727 Rs.

Question:8

The mean of the following frequency distribution is 50, but the frequencies f1 and f2 in classes 20-40 and 60-80, respectively are not known. Find these frequencies, if the sum of all the frequencies is 120.

Class

0-20

20-40

40-60

60-80

80-100

Frequency

17

fi

32

f2

19

Answer:

Solution.

Class

(fi)

xi

\mu _{c}= \frac{\left ( x_{i}-a \right )}{h}

fi\mu _{i}

0-20

17

10

-2

-34

20-40

f1

30

-1

-f1

40-60

32

50=a

0

0

60-80

f2

70

1

f2

80-100

19

90

2

38



\sum f_{i}= 68+f_{i}+f_{2}







Sum of all frequencies = 120
\Rightarrow68 + f1 + f2 = 120
f1 + f2 = 52 …(1)
a = 50, h = 20
mean = a+\frac{\sum f_{i}{\mu _{i}}}{\sum f_{i}}\times h
50= 50 + \frac{\left ( 4+f_{2}-f_{1} \right )\times 20}{20}
0= (4 + f2 – f1)
–f2 + f1 = 4 …(2)
add (1) and (2) we get
2f1 = 56 \Rightarrow f_{1}= 28
Put f1 = 28 in equation (1)
f2 = 52 – 28 \Rightarrow f_{2}= 24

Question:9

The median of the following data is 50. Find the values of p and q, if the sum of all the frequencies is 90.

Marks

Frequency

20-30
30-40
40-50
50-60
60-70
70-80
80-90

p
15
25
20
q
8
10

Answer:

Solution.

marks

Frequency

Cummulative frequency

20-30

1

p

30-40

15

15+p

40-50

25

40+p = cf

50-60

20=f

60+p

60-70

q

68+p+q

70-80

8

68+p+q

80-90

10

78+p+q

n = 90, \frac{n}{2}= 45
l = 50, f = 20, cf = 40 + p, h = 10
median = l+\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h
50= 50+\frac{\left ( 45-40-p \right )}{20}\times 10
0= \frac{5-p}{2}
5 – p = 0
p = 5
78 + 5 + q = 90
q = 90 – 83
q = 7

Question:10

The distribution of heights (in cm) of 96 children is given below :


Height (in cm)

Number of children

124-128
128-132
132-136
136-140
140-144
144-148
148-152
152-156
156-160
160-164

5
8
17
24
16
12
6
4
3
1

Draw a less than type cumulative frequency curve for this data and use it to compute median height of the children.

Answer:

Answer. [139]
Solution.

Height

Number of children

less than 124
less than 128
less than 132
less than 136
less than 140
less than 144
less than 148
less than 152
less than 156
less than 160
less than 164

0
5
13
30
54
70
82
88
92
95
96


\frac{n}{2}= \frac{96}{2}= 48
Hence the median is = 139

Question:11

Size of agricultural holdings in a survey of 200 families is given in the following table:

Size of agricultural holdings (in ha)

Number of families

0-5
5-10
10-15
15-20
20-25
25-30
30-35

10
15
30
80
40
20
5

Compute median and mode size of the holdings

Answer:

Answer. [17.77]
Solution.
mode= \iota +\left [ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \right ]\times h

Size of agricultural holdings

fi

cf

0-5

10

10

5-10

15

25

10-15

30

55

15-20

80

135

20-25

40

175

25-30

20

195

30-35

5

200

(i) Here n = 200
\frac{n}{2}= \frac{200}{2}= 100 which lies in interval (15 – 20)
l = 15, h = 5, f = 80 and cf = 55
median = l+\frac{\left ( \frac{n}{2}-cf \right )}{4}\times h= 15+\frac{\left ( 100-55 \right )\times 5}{80}
=15+\frac{45}{16}= 15+2\cdot 81= 17\cdot 81
l = 15, fm = 80, f1 = 30, f2 = 40 and h = 5
mode= \iota +\left [ \frac{f_{m}-f_{1}}{2f_{m}-f_{1}-f_{2}} \right ]\times h
= 15+\left [ \frac{80-30}{2\times 80-30-40} \right ]\times 5
= 15+\left [ \frac{50}{160-70} \right ]\times 5
= 15+\left [ \frac{50}{90} \right ]\times 5
=15+\frac{25}{9}
=15 + 2.77 = 17.77

Question:12

The annual rainfall record of a city for 66 days is given in the following table.

Rainfall (in cm)

0-10

10-20

20-30

30-40

40-50

50-60

Number of days

22

10

8

15

5

6

Calculate the median rainfall using ogives (of more than type and of less than type)

Answer:

Answer. [20]
Solution.

(i) less than type

(ii) more than type

Rain fall

No.of days

Rain fall

Number of days

less than 0

0

more than or equal to 0

66

less than 10

0+22 = 22

more than or equal to 10

66-22 = 44

less than 20

22+10 = 32

more than or equal to 20

44-10 = 34

less than 30

32+8 = 40

more than or equal to 30

34-8 = 26

less than 40

40+15 = 55

more than or equal to 40

26-15 = 11

less than 50

55+5 =60

more than or equal to 50

11 - 5 =6

less than 60

60+6 =66

more than or equal to 60

6-6 =0

Now let us draw ogives of more than type and of less than type then find the median

N= 66,\frac{N}{2}= \frac{66}{2}= 33
Here median is 20

Question:13

The following is the frequency distribution of duration for100 calls made on a mobile phone :

Duration (in seconds)

Number of calls

95-125
125-155
155-185
185-215
215-245

14
22
28
21
15

Calculate the average duration (in sec) of a call and also find the median from cumulative frequency curve.

Answer:

Answer. [170]
Solution.

Duration

fi

xi

u_{i}\frac{\left ( x_{i}-a \right )}{h}

fiui

95-125

14

110

-2

-28

125-155

22

140

-1

-22

155=185

28

170 = a

0

0

185-215

21

200

1

21

215-245

21

230

2

30



\sum f_{i}= 100





\sum f_{i}u_{i}= 1

a = 170, h = 30
Average = a+\frac{\sum f_{i}u_{i}}{\sum f_{i}}\times h= 170+\frac{1}{100}\times 30= 170+0\cdot 3= 170\cdot 3

less than type

Duration

Number of calls

less than 95
less than 125
less than 155
less than 185
less than 215
less than 245

0
0+14 = 14
14+22 =36
36 + 28 = 64
64 + 21 = 85
85 + 15 =100


n = 100
\frac{n}{2}= \frac{100}{2}= 50
median is 170

Question:14

50 students enter for a school javelin throw competition. The distance (in metres) thrown are recorded below :

Distance (in m)

0-20

20-40

40-60

60-80

80-100

Number of students

6

11

17

12

4

(i) Construct a cumulative frequency table.
(ii) Draw a cumulative frequency curve (less than type) and calculate the median distance thrown by using this curve.
(iii) Calculate the median distance by using the formula for median.
(iv) Are the median distance calculated in (ii) and (iii) same ?

Answer:

(i) Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables.

Distance

fi

CF

0-20

6

6

20-40

11

6+11 = 17

40-60

17

17+17 = 34

60-80

12

34 + 12 = 46

80-100

4

46 + 4 =50

(ii) Answer. [49.41]
Solution. Frequency distribution: It tells how frequencies are distributed over values in a frequency distribution. However mostly we use frequency distribution to summarize categorical variables


Distance

Cumulative frequency (C.F)

0
less than 20
less than 40
less than 60
less than 80
less than 100

0
6
17
34
46
50

n = 50
\frac{n}{2}= \frac{50}{2}= 25

median = 49.41

(iii)Answer. [49.41]
Solution. n = 50
\frac{n}{2}= \frac{50}{2}= 25

which lies in interval 40 – 60
l = 40, h = 20, CF = 17 and f = 17
median = l+\frac{\left ( \frac{n}{2}-cf \right )}{f}\times h
=40+\frac{\left ( 25-17 \right )}{17}\times 20
=40+\frac{8\times 20}{17}
=40 + 9.41
= 49.41

(iv) Yes, the median distance calculated in (ii) and (iii) are same.

NCERT Exemplar Solutions Class 10 Maths Chapter 13 Important Topics:

NCERT exemplar Class 10 Maths solutions chapter 13 deals with a wide range of concepts mentioned below:

  • How to find out the central tendency is of any given data.
  • Methods to find out the mean of any group data.
  • The direct method, step deviation method and assumed mean method to find out the mean of any group data.
  • NCERT exemplar Class 10 Maths solutions chapter 13 discusses cumulative frequency distribution and its graphical representation.
  • How to find out the mode and median of the given data.
  • Different techniques to find out medians.
  • Conditional probabilities, independent events, and Bayes theorem.

NCERT Class 10 Solutions for Other Subjects:

NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 13:

These Class 10 Maths NCERT exemplar chapter 13 solutions emphasise the methods to find out mean, median, and mode. In this chapter, students will understand the experimental and statistical approach of probability. Students will learn the condition for multiplying probability to find out the probability of any composite event. Statistics and Probability based practice problems can be easily studied and practiced using these Class 10 Maths NCERT exemplar solutions chapter 13 Statistics and Probability.

The students can comfortably sail through the NCERT Class 10 Maths, RD Sharma Class 10 Maths, A textbook for Mathematics by Monica Kapoor, and RS Aggarwal Class 10 Maths et cetera.

Check Chapter-Wise Solutions of Book Questions

Must, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Question (FAQs)

1. Make a List of topics and sub-topics covered in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths.

Below is a list of topics and subtopics covered in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths:

  1. Direct Method, Assumed Mean Method, and Step-Deviation Method for determining the mean of grouped data.
  2. Finding the mode of the given data.
  3. Calculating the median of grouped data.
  4. Graphical representation of cumulative frequency distribution.
2. In Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths, can you elucidate the concept of mean?

The mean is the arithmetic average of a given set of values, signifying an equal distribution of values in the dataset. Central tendency refers to the statistical measure that identifies a single value to represent the entire distribution, providing an accurate description of the whole data. This value is unique and represents the collected data. Mean, median, and mode are the three frequently used measures of central tendency.

3. What makes NCERT Exemplar Solutions for Class 10 Maths Chapter 13 advantageous for students to secure good grades in the board exam?

The subject experts at CAreers360 have developed the NCERT Exemplar Solutions for Chapter 13 of Class 10 Maths, keeping in mind the learning abilities of students. The solution PDF module is downloadable from BYJU'S website according to the students' needs. All crucial concepts are explained in plain language to aid students in achieving exam success with confidence. The solutions cover all problems in the NCERT textbook, allowing students to cross-check their answers and identify their areas of weakness.

4. Are these solutions accessible offline?

Yes, the NCERT exemplar Class 10 Maths solutions chapter 13 pdf download feature provided this solution for students practicing NCERT exemplar Class 10 Maths chapter 13.

5. List out the topics and sub-topics covered in Chapter 13 NCERT Exemplar Solutions for Class 10 Maths.

Below is a list of the topics and sub-topics included in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths:

  1. Calculation of mean for grouped data using direct method, assumed mean method and step-deviation method
  2. Determination of mode for the given dataset
  3. Computation of median for grouped data
  4. Graphical representation of cumulative frequency distribution.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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3 Jobs Available
Underwriter

An underwriter is a person who assesses and evaluates the risk of insurance in his or her field like mortgage, loan, health policy, investment, and so on and so forth. The underwriter career path does involve risks as analysing the risks means finding out if there is a way for the insurance underwriter jobs to recover the money from its clients. If the risk turns out to be too much for the company then in the future it is an underwriter who will be held accountable for it. Therefore, one must carry out his or her job with a lot of attention and diligence.

3 Jobs Available
Finance Executive
3 Jobs Available
Operations Manager

Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.

3 Jobs Available
Investment Director

An investment director is a person who helps corporations and individuals manage their finances. They can help them develop a strategy to achieve their goals, including paying off debts and investing in the future. In addition, he or she can help individuals make informed decisions.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
Transportation Planner

A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.

3 Jobs Available
Plumber

An expert in plumbing is aware of building regulations and safety standards and works to make sure these standards are upheld. Testing pipes for leakage using air pressure and other gauges, and also the ability to construct new pipe systems by cutting, fitting, measuring and threading pipes are some of the other more involved aspects of plumbing. Individuals in the plumber career path are self-employed or work for a small business employing less than ten people, though some might find working for larger entities or the government more desirable.

2 Jobs Available
Construction Manager

Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.

2 Jobs Available
Urban Planner

Urban Planning careers revolve around the idea of developing a plan to use the land optimally, without affecting the environment. Urban planning jobs are offered to those candidates who are skilled in making the right use of land to distribute the growing population, to create various communities. 

Urban planning careers come with the opportunity to make changes to the existing cities and towns. They identify various community needs and make short and long-term plans accordingly.

2 Jobs Available
Highway Engineer

Highway Engineer Job Description: A Highway Engineer is a civil engineer who specialises in planning and building thousands of miles of roads that support connectivity and allow transportation across the country. He or she ensures that traffic management schemes are effectively planned concerning economic sustainability and successful implementation.

2 Jobs Available
Environmental Engineer

Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems. 

2 Jobs Available
Naval Architect

A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability. 

2 Jobs Available
Orthotist and Prosthetist

Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.

6 Jobs Available
Veterinary Doctor
5 Jobs Available
Pathologist

A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.

5 Jobs Available
Speech Therapist
4 Jobs Available
Gynaecologist

Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth. 

4 Jobs Available
Oncologist

An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.

3 Jobs Available
Audiologist

The audiologist career involves audiology professionals who are responsible to treat hearing loss and proactively preventing the relevant damage. Individuals who opt for a career as an audiologist use various testing strategies with the aim to determine if someone has a normal sensitivity to sounds or not. After the identification of hearing loss, a hearing doctor is required to determine which sections of the hearing are affected, to what extent they are affected, and where the wound causing the hearing loss is found. As soon as the hearing loss is identified, the patients are provided with recommendations for interventions and rehabilitation such as hearing aids, cochlear implants, and appropriate medical referrals. While audiology is a branch of science that studies and researches hearing, balance, and related disorders.

3 Jobs Available
Hospital Administrator

The hospital Administrator is in charge of organising and supervising the daily operations of medical services and facilities. This organising includes managing of organisation’s staff and its members in service, budgets, service reports, departmental reporting and taking reminders of patient care and services.

2 Jobs Available
Actor

For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs. 

4 Jobs Available
Acrobat

Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.

3 Jobs Available
Video Game Designer

Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages.

Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.

3 Jobs Available
Radio Jockey

Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.

A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.

3 Jobs Available
Choreographer

The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.

2 Jobs Available
Videographer
2 Jobs Available
Multimedia Specialist

A multimedia specialist is a media professional who creates, audio, videos, graphic image files, computer animations for multimedia applications. He or she is responsible for planning, producing, and maintaining websites and applications. 

2 Jobs Available
Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook. 

5 Jobs Available
Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. 

Ever since internet costs got reduced the viewership for these types of content has increased on a large scale. Therefore, a career as a vlogger has a lot to offer. If you want to know more about the Vlogger eligibility, roles and responsibilities then continue reading the article. 

3 Jobs Available
Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available
Linguist

Linguistic meaning is related to language or Linguistics which is the study of languages. A career as a linguistic meaning, a profession that is based on the scientific study of language, and it's a very broad field with many specialities. Famous linguists work in academia, researching and teaching different areas of language, such as phonetics (sounds), syntax (word order) and semantics (meaning). 

Other researchers focus on specialities like computational linguistics, which seeks to better match human and computer language capacities, or applied linguistics, which is concerned with improving language education. Still, others work as language experts for the government, advertising companies, dictionary publishers and various other private enterprises. Some might work from home as freelance linguists. Philologist, phonologist, and dialectician are some of Linguist synonym. Linguists can study French, German, Italian

2 Jobs Available
Public Relation Executive
2 Jobs Available
Travel Journalist

The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.

2 Jobs Available
Welding Engineer

Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues. 

5 Jobs Available
QA Manager
4 Jobs Available
Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product. 

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Production Manager
3 Jobs Available
Merchandiser
2 Jobs Available
QA Lead

A QA Lead is in charge of the QA Team. The role of QA Lead comes with the responsibility of assessing services and products in order to determine that he or she meets the quality standards. He or she develops, implements and manages test plans. 

2 Jobs Available
Metallurgical Engineer

A metallurgical engineer is a professional who studies and produces materials that bring power to our world. He or she extracts metals from ores and rocks and transforms them into alloys, high-purity metals and other materials used in developing infrastructure, transportation and healthcare equipment. 

2 Jobs Available
Azure Administrator

An Azure Administrator is a professional responsible for implementing, monitoring, and maintaining Azure Solutions. He or she manages cloud infrastructure service instances and various cloud servers as well as sets up public and private cloud systems. 

4 Jobs Available
AWS Solution Architect

An AWS Solution Architect is someone who specializes in developing and implementing cloud computing systems. He or she has a good understanding of the various aspects of cloud computing and can confidently deploy and manage their systems. He or she troubleshoots the issues and evaluates the risk from the third party. 

4 Jobs Available
QA Manager
4 Jobs Available
Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
ITSM Manager
3 Jobs Available
Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.  

3 Jobs Available
Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack 

3 Jobs Available
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