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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 intelligencerelated 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 diâ€™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 d_{i} = x_{i} â€“ a
where x_{i} is data and a is mean
So, di are the derivative from midpoint 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 x_{i}â€™s are the mid points of the class intervals of grouped data, fiâ€™s are the corresponding frequencies and is the mean, then is equal to
(A) 0 (B) â€“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, u_{i} =
(A) (B) h(xi â€“ 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 d_{i} = x_{i} â€“ a and
put d_{i} = x_{i} â€“ 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  05  510  1015  1520  2025 
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) 
05  10  10 
510  15  (10+15=25) 
1015  12  (25+12=37) 
1520  20  (37+20=57) 
2025  9  (57+9=66) 
N=66 
which ies in the class 1015.
Hence the median class is 10 â€“ 15
The class with maximum frequency is modal class which is 15 â€“ 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  05  611  1217  1823  2429 
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 
05.5  13  13 
5.511.5  10  (13+10=23) 
11.517.5  15  (23+15=38) 
17.523.5  8  (38+8=46) 
23.529.5  11  (46+11=57) 
N = 57 
Here the median class is 15.5 â€“ 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) 1020 (B) 2030 (C) 3040 (D) 5060
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 
010  3  3 
1020  123=9  12 
2030  2712=15  27 
3040  5727=30  57 
4050  7557=18  75 
5060  8075=5  80 
=80 
The class with highest frequency is 3040
Hence 30 â€“ 40 is the modal class.
Question:9
Class  6585  85105  105125  125145  145165  165185  185205 
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 
6585  4  4 
85105  5  (4+5=9) 
105125  13  (9+13=22) 
125145  20  (22+20=42) 
145165  14  (42+14=56) 
165185  7  (56+7=63) 
185205  4  (63+4=67) 
N = 67 
Median class = 125 â€“ 145
upper limit of median = 145
The class with maximum frequency is modal class which is 125 â€“ 145
lower limit of modal class = 125
Difference of the upper limit of median and lower limit of modal = 145 â€“ 125 = 20
Question:10
The times, in seconds, taken by 150 athletes to run a 110 m hurdle race are tabulated below :
Class  13.814  1414.2  14.214.4  14.414.6  14.614.8  14.815 
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 (13.814) + frequency of class (14 14.2) +
frequency of class (14.214.4) + frequency of class (14.414.6)
= 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 3040 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 
010  63  5 
1020  58  3 
2030  55  4 
3040  51  3 
4050  48  6 
5060  42  42 
So the frequency of class 30 â€“ 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 â€“ 1 (B) p (C) 1 â€“ 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 â€“ 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 â‰¤ P(A) â‰¤ 1 (D) â€“1 â‰¤ P(A) â‰¤ 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 â‰¤ P(A) â‰¤ 1
It represent probability of event A because probability of an event is always lies from 0 to 1.
(D) â€“1 â‰¤ P(A) â‰¤ 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 â€“ 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 Ã— 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 â€“ 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 midpoint.
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 â€“ 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  13  35  57  710 
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 (x_{i})  Frequency(f_{i})  f_{ix}i 
13  2  9  18 
35  4  22  88 
57  6  27  162 
710  8.5  17  144.5 
Question:2
Calculate the mean of the scores of 20 students in a mathematics test :
Marks  1020  2030  3040  4050  5060 
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  x_{i}  No.of students fi  f_{ix}i 
1020  15  2  30 
2030  25  4  100 
3040  35  7  245 
4050  45  6  270 
5060  55  1  55 
Question:3
Calculate the mean of the following data
Class  47  811  1215  1619 
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  x_{i}  fi  f_{i x}i 
47  5.5  5  275 
811  9.5  4  38 
1215  13.5  9  121.5 
1619  17.5  10  175 
Question:4
no.of pages written per day  1618  1921  2224  2527  2830 
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)  f_{ix}i 
1618  1  17  17 
1921  3  20  60 
2224  4  23  92 
2527  9  26  234 
2830  13  29  377 
Question:5
The daily income of a sample of 50 employees are tabulated as follows :
Income (in Rs)  1200  201400  401600  601800 
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  f_{ix}i  
1200  100.5  14  1407  
201400  300.5  15  4507.5  
401600  500.5  14  7007  
601800  700.5  7  4903.5  
Question:6
no.of seats  100104  104108  108112  112116  116120 
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  f_{ix}i 
100104  15  102  1530 
104108  20  106  2120 
108112  32  110  3520 
112116  18  114  2052 
116120  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)  100110  110120  120130  130140  140150 
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  f_{ix}i 
100110  4  105  420 
110120  14  115  1610 
120130  21  125  2625 
130140  8  135  1080 
140150  3  145  435 
Question:8
MIleage (km/I)  1020  1214  1416  1618 
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  f_{ix}i 
1012  7  11  77 
1214  12  13  156 
1416  18  15  270 
1618  13  17  221 

Question:9
The following is the distribution of weights (in kg) of 40 persons :
Weight (in kg)  4045  4550  5055  5560  6065  6570  7075  7075 
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 
4045  4  4 
4550  4  8 
5055  13  21 
5560  5  26 
6065  6  32 
6570  5  37 
7075  2  39 
7580  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 
010  10  10 
1020  50  5010=40 
2030  130  13050=80 
3040  270  270130=140 
4050  440  440270=170 
5060  570  570440=130 
6070  670  670570=100 
7080  740  740670=70 
8090  780  780740=40 
90100  800  800780=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 
010  3432=2 
1020  3230=2 
2030  3027=3 
3040  2723=4 
4050  2317=6 
5060  1711=6 
6070  116=5 
7080  64=2 
8090  4 
Question:12
Height (in cm)  Frequency  Cumulative frequency 
150155  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 â€“ 12
b = 13
25 + 10 = c
35= c
c + d = 43
35 + d = 43
d = 43 â€“ 35
d=8
43 + e = 48
e = 48 â€“ 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)  1020  2030  3040  4050  5060  6070 
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  2217 = 5 
40 60  29  29 22 = 7 
60 80  37  3729 = 8 
80100  50  50 37 = 13 
Question:15
Weekly income of 600 families is tabulated below :
Weekly income (in Rs)  Number of families 
01000  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)  85100  100115  115130  130145 
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 
05000  8 
Calculate the modal income.
Answer:
Answer. [11875]
Solution. Here l = 10000, f_{1} = 41, f_{0} = 26, f_{2} = 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 
200201  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 â€“ same number case
= 36 â€“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 â€“ (blue ball)
= 22 â€“ 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 â€“ 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 â€“ 3 ( three cards are removed)
= 49
Total king = 4 â€“ 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 â€“ 3 = 49 ( three cards are removed)
Total club = 13 â€“ 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 â€“ 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 â€“ 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 â€“ 12 = 40 ( 12 cards are removed)
Cards greater than 7 =8,9,10 (3 Ã— 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 â€“ 12 = 40 ( 12 cards are removed)
Cards less than 7 = 1, 2, 3, 4, 5, 6 (6 Ã— 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 â€“ envelopes with cash prize
= 1000 â€“ 10 â€“ 100 â€“ 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, nondefective 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 â€“ 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 â€“ (in which she gets entry book + in which she gets double)
= 8 â€“ 6 (HHT, HTH, THH, TTH, THT, HTT) â€“ 1(HHH)
= 8 â€“ 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 Ã— 4 = 8
Blue balls = 3x = 3 Ã— 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 Ã— 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 â€“ 1 = 999
Now the total winning cards = 9 â€“ 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 
010  5  80  8077 = 3  15 
1020  15  77  78772 = 5  75 
2030  25  72  7265 =7  175 
3040  35  65  6555 = 10  350 
4050  45  55  5543 = 12  540 
5060  55  43  4328 = 15  825 
6070  65  28  2816 = 12  780 
7080  75  16  1610 =6  450 
8090  85  10  108 = 2  170 
90100  95  8  80 = 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 
010  5  5  5  15 
1020  15  9  95 =4  75 
2030  25  17  179 = 8  175 
3040  35  29  2917 = 12  350 
4050  45  45  4529 = 16  540 
5060  55  60  6045 = 15  825 
6070  65  70  7060 = 10  780 
7080  75  78  7870 = 8  450 
8090  85  83  8378 = 5  170 
90100  95  85  8583 = 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  
010  5  10090 =10  50 
1020  15  9075 = 15  225 
2030  25  7550 = 25  625 
3040  35  5025 =25  875 
4050  45  2515 =10  450 
5060  55  155 = 10  550 
6070  65  50 = 5  325 
Question:4
The weights of tea in 70 packets are shown in the following table :
Weight (in gram)  Number of packets 
200201  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 
200201  200.5  13  2606.5 
201202  201.5  27  5440.5 
202203  202.5  18  3645.0 
203204  203.5  10  2035.0 
204205  204.5  1  204.5 
205206  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 
200201  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 
200201  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  7013 = 57 
Less than 202  40  More than or equal to 202  5727 =30 
Less than 203  58  More than or equal to 203  3018 =12 
Less than 204  68  More than or equal to 204  1210 = 2 
Less than 205  69  More than or equal to 205  21 = 1 
Less than 206  70  More than or equal to 206  11 = 0 
Hence median = 201.8
Question:7
The table below shows the salaries of 280 persons.
Salary(in thousand (Rs))  Number of persons 
510  49 
Calculate the median and mode of the data.
Answer:
Solution.
Salary  fi  cf 
510  49  49 
1015  133  49+133=182 
1520  63  182+63=245 
2025  15  245+15 = 260 
2530  6  260+6 = 266 
3035  7  266+7 = 273 
3540  4  273+4 = 277 
4045  2  277+2 = 279 
4550  1  279+1 = 280 
f_{1} = 49, f_{m}= 133, f_{2}= 63, cf = 49, f = 133
l = 10, h = 5
median =
=
=
=
In thousands = 13.421 Ã— 1000 = 13421 Rs.
Mode =
=
=
=10 + 2.727
=12.727
In thousands = 12.727 Ã— 1000 = 12727 Rs.
Question:8
Class  020  2040  4060  6080  80100 
Frequency  17  f_{i}  32  f_{2}  19 
Answer:
Solution.
Class  (f_{i})  x_{i}  f_{i}  
020  17  10  2  34 
2040  f_{1}  30  1  f_{1} 
4060  32  50=a  0  0 
6080  f_{2}  70  1  f_{2} 
80100  19  90  2  38 
Sum of all frequencies = 120
68 + f_{1} + f_{2} = 120
f_{1} + f_{2} = 52 â€¦(1)
a = 50, h = 20
mean =
50= 50 +
0= (4 + f_{2} â€“ f_{1})
â€“f_{2} + f_{1} = 4 â€¦(2)
add (1) and (2) we get
2f_{1} = 56
Put f_{1} = 28 in equation (1)
f_{2} = 52 â€“ 28
Question:9
Marks  Frequency 
2030  p 
Answer:
Solution.
marks  Frequency  Cummulative frequency 
2030  1  p 
3040  15  15+p 
4050  25  40+p = cf 
5060  20=f  60+p 
6070  q  68+p+q 
7080  8  68+p+q 
8090  10  78+p+q 
n = 90,
l = 50, f = 20, cf = 40 + p, h = 10
median =
5 â€“ p = 0
p = 5
78 + 5 + q = 90
q = 90 â€“ 83
q = 7
Question:10
The distribution of heights (in cm) of 96 children is given below :
Height (in cm)  Number of children 
124128  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 
05  10 
Compute median and mode size of the holdings
Answer:
Answer. [17.77]
Solution.
Size of agricultural holdings  fi  cf 
05  10  10 
510  15  25 
1015  30  55 
1520  80  135 
2025  40  175 
2530  20  195 
3035  5  200 
(i) Here n = 200
which lies in interval (15 â€“ 20)
l = 15, h = 5, f = 80 and cf = 55
=
l = 15, f_{m} = 80, f_{1} = 30, f_{2} = 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)  010  1020  2030  3040  4050  5060 
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  6622 = 44 
less than 20  22+10 = 32  more than or equal to 20  4410 = 34 
less than 30  32+8 = 40  more than or equal to 30  348 = 26 
less than 40  40+15 = 55  more than or equal to 40  2615 = 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  66 =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 
95125  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  f_{i}u_{i}  
95125  14  110  2  28 
125155  22  140  1  22 
155=185  28  170 = a  0  0 
185215  21  200  1  21 
215245  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)  020  2040  4060  6080  80100 
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  f_{i}  CF 
020  6  6 
2040  11  6+11 = 17 
4060  17  17+17 = 34 
6080  12  34 + 12 = 46 
80100  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 â€“ 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 crosscheck 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 subtopics included in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths:
Application Date:20 November,2023  19 December,2023
Application Date:20 November,2023  19 December,2023
The eligibility age criteria for class 10th CBSE is 14 years of age. Since your son will be 15 years of age in 2024, he will be eligible to give the exam.
According to CBSE norms, a student must be 14 years old by the end of the year in which the exam will be held in order to sit for the 10th board exam. If your age is greater than 14, however, there are no limits. Therefore, based on your DOB, you will be 12 years and 6 months old by the end of December 2022. After the actual 16 June 2024, you'll be qualified to take your 10th board.
Hello aspirant,
Central Board Of Secondary Education(CBSE) is likely to declare class 10 and 12 terms 2 board result 2022 by July 15. The evaluation process is underway. Students are demanding good results and don't want to lack behind. They are requesting the board to use their best scores in Term 1 and Term 2 exams to prepare for the results.
CBSE concluded board exams 2022 for 10, and 12 on June 15 and May 24. Exams for both classes began on April 26. A total of 35 lakh students including 21 lakh class 10 students and 14 lakh class 12 students appeared in exams and are awaiting their results.
You can look for your results on websites cbse.gov (//cbse.gov) .in, cbseresults.nic.in
Thank you
Hello SIR CBSE board 9th class admission 2022 Kara Raha hu entrance ki problem hai please show the Entrance for Baal vidya mandir school sambhal
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Quality Assurance Manager Job Description: A QA Manager is an administrative professional responsible for overseeing the activity of the QA department and staff. It involves developing, implementing and maintaining a system that is qualified and reliable for testing to meet specifications of products of organisations as well as development processes.
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.
Are you searching for a Reliability Engineer job description? A Reliability Engineer is responsible for ensuring long lasting and high quality products. He or she ensures that materials, manufacturing equipment, components and processes are error free. A Reliability Engineer role comes with the responsibility of minimising risks and effectiveness of processes and equipment.
A Safety Manager is a professional responsible for employee’s safety at work. He or she plans, implements and oversees the company’s employee safety. A Safety Manager ensures compliance and adherence to Occupational Health and Safety (OHS) guidelines.
Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.
ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.
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
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 selftaught 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, wordprocessing 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.
An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.
A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.
The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.
A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.
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