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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.
In the formula
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 is the mean, then 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
By cross multiplication we get
(from equation (1))
= 0
Question:4
In the formula for finding the mean of grouped frequency distribution, ui =
(A) (B) h(xi – a) (C) (D)
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
put di = xi – a
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 |
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 |
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 | 3 |
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
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 |
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 |
|
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) (C) (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 =
Probability =
Question:13
Which of the following cannot be the probability of an event?
(A) (B) 0.1 (C) 3% (D)
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
Hence 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)
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) (B) (C) (D)
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)
Question:19
The probability that a non leap year selected at random will contain 53 Sundays is
(A) (B) (C) (D)
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 =
Question:20
When a die is thrown, the probability of getting an odd number less than 3 is
(A) (B) (C) (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 =
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) =
0.035 =
Number of favourable cases =
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)
0.08 × 6000 = Number of tickets the bought
= Number of tickets the bought
Number of tickets the bought = 480.
Question:24
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) =
Question:25
Someone is asked to take a number from 1 to 100. The probability that it is a prime is
(A) (B) (C) (D)
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 =
Question:26
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
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
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
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
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 =
Probability of 0 girl =
Probability of 1 girl =
Probability of 2 girl =
Probability of 3 girl =
Here they are not equal to
Question:6
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.
All probabilities are not equal. So the given statement is false.
Question:7
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 =
Probability that Apoorv get 36 =
Peehu throws are die total cases = 6
Square of 6 is 36
Hence case = 1
Probability that Peehu get 36 =
Hence Peehu has better cases to get 36.
Question:8
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 =
Probability of head =
Probability of tail =
Hence the probability of each outcome is .
Question:9
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 =
Probability of getting
Number of favourable cases 'not 1' = 5 (2, 3, 4, 5, 6)
Probability of not 1 =
Hence they are not equal to
Question:10
Answer:
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 cases in tossing three coins = 8(HHH, HHT, HTH, THH, TTT, TTH, THT, HTT)
Number of case with no head = TTT
Probability =
Probability of no head =
The conclusion that probability of no head is is wrong because as we calculate it above, it comes out . Hence the probability of no head is
Question:11
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
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 .
So, it is not necessary that she gets tail at fourth toss. She can get head also.
Question:13
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 .
Hence tail is not have higher chance than head.
Both are have equal chance.
Question:14
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 =
Probability of slip with even number =
Slips with odd number = 50
Probability of slip with odd number =
Hence the probability of each is .
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:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
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:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
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 |
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:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
Question:4
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:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
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:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 | |
Question:6
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:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
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:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
Question:8
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 |
|
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
Marks | Number of students |
Below 10 | 10 |
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 | 4 |
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
Height (in cm) | Frequency | Cumulative frequency |
150-155 | 12 | a |
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.
( 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
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 | 60+42+55+70+53+20 = 300 |
Question:14
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 | 250 |
Total | 600 |
Compute the median income.
Answer:
Answer. [1263.15]
Solution. n = 600
l= 1000, l = 1000, cf = 250, f = 190
median =
Median = 1263.15
Question:16
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
h = 15, f = 9 , cf = 11
Median
= 100 + 9.16 109.16
Question:17
The monthly income of 100 families are given as below :
Income (in Rs) | Number of families |
0-5000 | 8 |
Calculate the modal income.
Answer:
Answer. [11875]
Solution. Here l = 10000, f1 = 41, f0 = 26, f2 = 16, h = 5000
Mode =
=
=
= 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 | 12 |
Determine the modal weight.
Answer:
Answer. [201.7 g]
Solution.
Here l = 201, f1 = 26, f0 = 12, f2 = 20, h = 1
Question:19
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)] =
(ii) Different number cases = 36 – same number case
= 36 –6 = 30
Let A be the event of getting different number
Probability [p(A)]=
Question:20
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)]=
Probability of getting sum 7 =
(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 =
Probability that sum is a prime number =
(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 =
Probability of getting sum 1 =
Question:21
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 =
Probability of getting product 6 =
(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 =
Probability of getting product 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
Product 7 = 0 (case)
Cases = 0
Probability =
Probability of getting product 7 =
Question:22
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 =
Probability of getting product less than 9 =
Question:23
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 =
Probability of getting sum 2 =
case of getting sum 3 = (1, 2), (1, 2), (2, 1), (2, 1)
Probability =
Probability of getting sum 3=
case of getting sum 4 = (1, 3), (1, 3), (2, 2), (2, 2), (3, 1), (3, 1)
Probability =
Probability of getting sum 4=
case of getting sum 5 = (2, 3), (2, 3), (4, 1),(4,1) (3, 2), (3, 2)
Probability =
Probability of getting sum 5 =
case of getting sum 6 = (3, 3), (3, 3), (4, 2), (4, 2), (5, 1), (5, 1)
probability =
Probability of getting sum 6=
case of getting sum 7 = (4, 3), (4, 3), (5, 2), (5, 2), (6, 1), (6, 1)
probability =
Probability of getting sum 7=
case of getting sum 8 = (5, 3), (5, 3), (6, 2), (6, 2)
probability =
Probability of getting sum 8=
case of getting sum 9 = (6, 3), (6, 3)
probability =
Probability of getting sum 9=
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 =
Probability of getting at most 1 head =
Question:25
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 =
Probability of getting all heads =
(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 =
Probability of getting at least 2 heads =
Question:26
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 =
Probability of getting difference 2 =
Question:27
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 =
Probability of getting red ball =
(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 =
Probability of getting green ball =
(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 =
Probability of getting not a blue ball =
Question:28
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 =
Probability of getting a heart =
(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 =
Probability of getting a King=
Question:29
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 =
Probability of getting a club =
(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 =
Probability of getting a heart =
Question:30
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 =
Probability of getting card 7=
(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 ( 12 cards are removed)
Cards greater than 7 =8,9,10 (3 × 4 = 12)
favourable cases = 12
probability =
Probability of getting card 7=
(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 ( 12 cards are removed)
Cards less than 7 = 1, 2, 3, 4, 5, 6 (6 × 4 = 24)
favourable cases = 24
probability =
Probability of getting card 7=
Question:31
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 =
Probability of getting number divisible by 7 =
(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 =
Probability of getting number divisible by 7 =
Question:32
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 =
Probability that card is with even number =
(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 =
Probability that the card is with a square number =
Question:33
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 =
Probability that alphabet is consonant =
Question:34
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 =
Probability that the envelope is no cash prize =
Question:35
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 =
Probability that slips is not marked 1
Question:36
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 =
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 =
Now probability that the bulb is defective = .
Question:37
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 =
Probability that piece is a triangle
(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 =
Probability that the piece is a square =
(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 =
Probability that piece is a square of blue color =
(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 =
Probability that piece is a triangle of red colour
Question:38
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 =
Probability that she will lose money =
(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 =
Probability that she gets double entry fee =
(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 =
Probability that she gets entry fees =
Question:39
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 =
Probability of getting a total 7 =
Question:40
(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 =
Probability that acceptable to Varnika =
(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 =
Probability that trader accept
Question:41
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 =
Probability that ball is not red =
(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 =
Probability of getting on white ball =
Question:42
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 =
Probability that player wins =
(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 =
Probability that second player wins after first =
Question:1
Find the mean marks of students for the following distribution
Marks | Number of students |
0 and above | 80 |
Answer:
Answer. [51.75]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
Question:2
Determine the mean of the following distribution :
Marks | Number of students |
Below 10 | 5 |
Answer:
Answer. [48.4]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
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:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
Question:4
The weights of tea in 70 packets are shown in the following table :
Weight (in gram) | Number of packets |
200-201 | 13 |
Find the mean weight of packets.
Answer:
Answer. [201.95]
Solution. Here we calculate mean by following the given steps:
Find the mid point of each interval.
Multiply the frequency of each interval by its mid point.
Get the sum of all the frequencies (f) and sum of all the (fx)
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 |
Question:5
The weights of tea in 70 packets are shown in the following table :
Weight (in gram) | Number of packets |
200-201 | 13 |
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 |
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 | 13 |
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 |
Hence median = 201.8
Question:7
The table below shows the salaries of 280 persons.
Salary(in thousand (Rs)) | Number of persons |
5-10 | 49 |
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 |
f1 = 49, fm= 133, f2= 63, cf = 49, f = 133
l = 10, h = 5
median =
=
=
=
In thousands = 13.421 × 1000 = 13421 Rs.
Mode =
=
=
=10 + 2.727
=12.727
In thousands = 12.727 × 1000 = 12727 Rs.
Question:8
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 |
Frequency | 17 | fi | 32 | f2 | 19 |
Answer:
Solution.
Class | (fi) | xi | fi | |
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 of all frequencies = 120
68 + f1 + f2 = 120
f1 + f2 = 52 …(1)
a = 50, h = 20
mean =
50= 50 +
0= (4 + f2 – f1)
–f2 + f1 = 4 …(2)
add (1) and (2) we get
2f1 = 56
Put f1 = 28 in equation (1)
f2 = 52 – 28
Question:9
Marks | Frequency |
20-30 | p |
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,
l = 50, f = 20, cf = 40 + p, h = 10
median =
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 | 5 |
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 | 0 |
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 | 10 |
Compute median and mode size of the holdings
Answer:
Answer. [17.77]
Solution.
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
which lies in interval (15 – 20)
l = 15, h = 5, f = 80 and cf = 55
=
l = 15, fm = 80, f1 = 30, f2 = 40 and h = 5
=
=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
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 | 14 |
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 | 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 |
a = 170, h = 30
Average =
less than type | |
Duration | Number of calls |
less than 95 | 0 |
n = 100
median is 170
Question:14
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 | 0 |
n = 50
median = 49.41
(iii)Answer. [49.41]
Solution. n = 50
which lies in interval 40 – 60
l = 40, h = 20, CF = 17 and f = 17
=
=
=40 + 9.41
= 49.41
(iv) Yes, the median distance calculated in (ii) and (iii) are same.
NCERT exemplar Class 10 Maths solutions chapter 13 deals with a wide range of concepts mentioned below:
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.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
Below is a list of topics and subtopics covered in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths:
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.
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.
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.
Below is a list of the topics and sub-topics included in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths:
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Thanking you
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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.
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.
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.
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.
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.
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