NCERT Exemplar Class 10 Maths Solutions for Chapter 13 - Statistics and Probability

NCERT Exemplar Solutions Class 10 Maths Chapter 13 – Free PDF Download

The NCERT Exemplar Class 10 Maths Chapter 13 Statistics and Probability is provided here for students to prepare for CBSE exams. These problems and solutions are prepared by our subject experts in accordance with CBSE’s latest syllabus (2023-2024) to help students score better. Students can use the NCERT Exemplar for Class 10 Maths for studying as well as practising questions. The exemplar also contains solved questions relevant to the exercise problems present in the textbook. Moreover, it will allow students to have a thorough revision of the entire chapter and be prepared to face the board exam.

Chapter 13 has problems based on the different statistical measures like mean, mode and median. Students will learn about how to solve these problems and also explains the concept of cumulative frequency distribution, cumulative frequency curves and more. Additionally, students will also develop an experimental approach to probability. Furthermore, they will explore concepts like the multiplication rule of probability, Bayes’ theorem, and the independence of events. These Exemplars problems and solutions cover the following topics in the chapter Statistics and Probability:

• Determining the mean of grouped data by a direct method, assumed mean method and step-deviation method
• Finding the mode of the given data
• To find the median of the grouped data
• Representation of cumulative frequency distribution graphically

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Exercise 13.1 Page No: 157

Choose the correct answer from the given four options:

1. In the formula

For finding the mean of grouped data di’s are deviations from a of

(A) Lower limits of the classes

(B) Upper limits of the classes

(C) Mid points of the classes

(D) Frequencies of the class marks

Solution:

(C) Mid points of the classes

Explanation:

We know,

d_i = x_i – a

Where,

x_i are data and ‘a’ is the assumed mean

So, d_i are the deviations from of mid – points of the classes.

Hence, the option (C) is correct

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

Solution:

(B) Centered at the class marks of the classes

Explanation:

In computing the mean of grouped data, the frequencies are centered at the class marks of the classes.

Hence, the option (B) is correct

3. If xi’s are the mid points of the class intervals of grouped data, fi’s are the corresponding frequencies and x is the mean, then (f_ix_i –

(A)0 (B) –1 (C) 1 (D) 2

Solution:

(A) 0

Explanation:

Mean (x) = Sum of all the observations/ Number of observations

x = (f₁x₁ + f₂x₂ + …..+ f_nx_n) / f₁ + f₂ +……+ f_n

x = Σfixi / Σfi, Σfi = n

x = Σfixi / n

n x = Σfixi …………… (1)

Σ (fixi – x) = (f₁x₁ – x) + (f₂x₂ – x)+ …..+ (fnxn – x)

Σ (fixi – x) = (f₁x₁ + f₂x₂ + …..+ fnxn) – (x +x +….n times)

Σ (fixi – x) = Σfixi –nx

Σ(fixi – x) = nx – nx ( From eq1)

Σ(fixi – x) = 0

Hence, option (A) is correct

4. In the formula x = a + h(f_iu_i/f_i), for finding the mean of grouped frequency distribution, u_i =

(A) (x_i+a)/h

(B) h (xi – a)

(C) (x_i –a)/h

(D) (a – x_i)/h

Solution:

(C) (x_i –a)/h

Explanation:

According to the question,

x = a + h(f_iu_i/f_i),

Above formula is a step deviation formula.

In the above formula,

x_i is data values,

a is assumed mean,

h is class size,

When class size is same we simplify the calculations of the mean by computing the coded mean of u₁,u₂,u₃…..,

Where u_i = (x_i – a)/h

Hence, option (C) is correct

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

Solution:

(B) Median

Explanation:

Since, the intersection point of less than ogive and more than ogive gives the median on the abscissa, 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

Hence, option (B) is correct

6. For the following distribution :

Class 0-05 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

Solution:

(B) 25

Explanation:

From the table, N/2 = 66/2 = 33, which lies in the interval 10 – 15.

Hence, lower limit of the median class is 10.

The highest frequency is 20, which lies in between the interval 15 – 20.

Hence, lower limit of modal class is 15.

Therefore, required sum is 10 + 15 = 25.

Hence, option (B) is correct

7. Consider the following frequency distribution:

Class 0-05 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

Solution:

(B) 17.5

Explanation:

According to the question,

Classes are not continuous, hence, we make the data continuous by subtracting 0.5 from lower limit and adding 0.5 to upper limit of each class.

According to the question,

N/2 = 57/2 = 28.5

28.5 lies in between the interval 11.5 – 17.5.

Therefore, the upper limit is 17.5.

Hence, option (B) is correct

8. For the following distribution:

Marks Number of students

Below 10 3

Below 20 12

Below 30 27

Below 40 57

Below 50 75

Below 60 80

The modal class is

(A)10-20 (B) 20-30 (C) 30-40 (D) 50-60

Solution:

(C) 30-40

Explanation:

Here, we see that the highest frequency is 30, which lies in the interval 30 – 40.

Hence, option (C) is correct

9. Consider the data :

The difference of the upper limit of the median class and the lower limit of the modal class is

A0 (B) 19 (C) 20 (D) 38

Solution:

(C) 20

Explanation:

Here, N/2 = 67/2 = 33.5 which lies in the interval 125 – 145.

Hence, upper limit of median class is 145.

Here, we see that the highest frequency is 20 which lies in 125 – 145.

Hence, the lower limit of modal class is 125.

∴ Required difference = Upper limit of median class – Lower limit of modal class

= 145 – 125 = 20

Hence, option (C) is correct

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

The number of athletes who completed the race in less than 14.6 seconds is :

A11 (B) 71 (C) 82 (D) 130

Solution:

(C) 82

Explanation:

The number of athletes who completed the race in less than 14.6 second= 2 + 4 + 5 + 71 = 82

Hence, option (C) is correct

11. Consider the following distribution :

Marks obtained Number of students

More than or equal to 0 63

More than or equal to 10 58

More than or equal to 20 55

More than or equal to 30 51

More than or equal to 40 48

More than or equal to 50 42

The frequency of the class 30-40 is

(A) 3 (B) 4 (C) 48 (D) 51

Solution:

(A) 3

Explanation:

Hence, frequency in the class interval 30 – 40 is 3.

Hence, option (A) is correct

12. If an event cannot occur, then its probability is

(A)1 (B) ¾ (C) ½ (D) 0

Solution:

(D) 0

Explanation:

The event which cannot occur is said to be impossible event.

The probability of impossible event = zero.

Hence, option (D) is correct

13. Which of the following cannot be the probability of an event?

(A)1/3 (B) 0.1 (C) 3% (D)17/16

Solution:

(D)17/16

Explanation:

Probability of an event always lies between 0 and 1.

Probability of any event cannot be more than 1 or negative as (17/16) > 1

Hence, option (D) is correct

Exercise 13.2 Page No: 161

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

Solution:

In order to calculate the median of a grouped data, the formula used is based on the assumption that the observations in the classes are uniformly distributed or equally spaced. Hence, we cannot say that the statement “the median of an ungrouped data and the median calculated when the same data is grouped are always the same” is always correct.

2. In calculating the mean of grouped data, grouped in classes of equal width, we may use the formula

where a is the assumed mean. a must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

Solution:

No, the statement is not correct. It is not necessary that assumed mean should be the mid – point of the class interval. a can be considered as any value which is easy to simplify it.

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

Solution:

No, the values of mean, mode and median of grouped data can be the same as well, it depends on the type of data given.

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

Solution:

The median class and modal class of grouped data is not always different, it depends on the data given.

5. In a family having three children, there may be no girl, one girl, two girls or three girls. So, the probability of each is ¼. Is this correct? Justify your answer.

Solution:

No it is not correct that in a family having three children, there may be no girl, one girl, two girls or three girls, the probability of each is ¼. .

Let boys be B and girls be G

Outcomes can be BBB , GGG , BBG , BGB , GBB, GGB, GBG , BGG

Then Probability of 3 girls = 1/8

Probability of 0 girls = 1/8

Probability of 2 girls = 3/8

Probability of 1 girl = 3/8

6. A game consists of spinning an arrow which comes to rest pointing at one of the regions (1, 2 or 3) (Fig. 13.1). Are the outcomes 1, 2 and 3 equally likely to occur? Give reasons.

Solution:

Total no. of outcome = 360

p(1)= 90/360 =1/4

p(2) = 90/ 360 = 1/4

p(3) = 180/360 =1/2

Hence, it is clear that the outcome are not equal

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

Solution:

Apoorv throw two dice at once.

Hence, the total number of outcomes = 36

Number of outcomes for getting product 36 = 1(6×6)

∴ Probability for Apoorv = 1/36

Peehu throws one die,

Hence, the total number of outcomes = 6

Number of outcomes for getting square = 36

∴ Probability for Peehu = 6/36 = 1/6

Therefore, Peehu has a better chance of getting the number 36.

Exercise 13.3 Page No: 166

1. Find the mean of the distribution :

Class 1-3 3-5 5-7 7-10

Frequency 9 22 27 17

Solution:

We first, find the class mark x_i of each class and then proceed as follows.

Mean,

Therefore, mean of the given distribution = 5.5.

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

Solution:

We first, find the class mark x_i of each class and then proceed as follows

Mean,

Therefore, mean of scores of 20 students in mathematics test = 35.

3. Calculate the mean of the following data :

Class 4 – 7 8 –11 12– 15 16 –19

Frequency 5 4 9 10

Solution:

The given data is not continuous.

So, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.

Mean,

Therefore, mean of the given data = 12.93.

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

Number of pages written per day 16-18 19-21 22-24 25-27 28-30

Number of days 1 3 4 9 13

Find the mean number of pages written per day.

Solution:

The given data is not continuous.

Hence, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.

Therefore, the mean of pages written per day = 26.

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 15 14 7

Find the mean daily income of employees.

Solution:

∴ Assumed mean, a = 300.5 and d_i = (x_i – a)

= 300.5 + 2800/50

= 356.5

Hence, the average daily income of employees = Rs.356.5

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

Number 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.

Solution:

∴ Assumed mean, a = 110

Class width, h = 4

And total observations, N = 100

Hence, finding mean,

= 110 + (-8/100)

= 110 – 0.08

= 109.92

But we know that the seats cannot be in decimal.

Therefore, the number of seats = 109.

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

Find the mean weight of the wrestlers.

Solution:

∴ Assumed mean, (a) = 125

Class width, (h) = 10

and total observations, (N) = 50

By step deviation method,

= 125 – 16

= 123.4kg

Hence, mean weight of wrestlers = 123.4kg

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

Mileage (km/l) 10-12 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?

Solution:

Here, Σf_i = 50

Σf_ix_i = 724

= 724/50 = 14.48

Hence, mean mileage = 14.48 km/h

No, I don’t agree with the claim because the manufacturer is claiming mileage 1.52 km/h more than average mileage.

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 75-80

Number of persons 4 4 13 5 6 5 2 1

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

Solution:

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

Marks Number of students

Below 10 10

Below 20 50

Below 30 130

Below 40 270

Below 50 440

Below 60 570

Below 70 670

Below 80 740

Below 90 780

Below 100 800

Construct a frequency distribution table for the data above.

Solution:

The frequency distribution table for the given data is:

11. Form the frequency distribution table from the following data :

Marks (out of 90) Number of candidates

More than or equal to 80 4

More than or equal to 70 6

More than or equal to 60 11

More than or equal to 50 17

More than or equal to 40 23

More than or equal to 30 27

More than or equal to 20 30

More than or equal to 10 32

More than or equal to 0 34

Solution:

The frequency distribution table for the given data is:

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

Height Frequency Cumulative frequency

(in cm)

150-155 12 a

155-160 b 25

160-165 10 c

165-170 d 43

170-175 e 48

175-180 2 f

Total 50

Solution:

On comparing last two tables, we get

a = 12

∴ 12 + b = 25

⇒ b = 25 – 12 = 13

22 + b = c

⇒ c = 22 + 13 = 35

22 + b + d = 43

⇒ 22 + 13 + d = 43

⇒ d = 43 – 35 = 8

22 + b + d + e = 48

⇒ 22 + 13 + 8 + e = 48

⇒ e = 48 – 43 = 5

24 + b + d + e = f

⇒ f = 24 + 13 + 8 + 5 = 50

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

Age (in years) 10-20 20-30 30-40 40-50 50-60 60-70

Number of patients 60 42 55 70 53 20

Form:

ALess than type cumulative frequency distribution.

• More than type cumulative frequency distribution.Solution:(i)Less than type cumulative frequency distribution of the data is given below.
  

  (i) Less than type
  Age (in year)	Number of patients
  Less than 10	0
  Less than 20	60 + 0 = 60
  Less than 30	60 + 42 = 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

  (ii)

  More than type cumulative frequency distribution of the data is given below.

  (i) More than type
  Age (in year)	Number of patients
  More than or equals 10	60 + 42 + 55 + 70 + 53 + 20 = 300
  More than or equals 20	42 + 55 + 70 + 53 + 20 = 240
  More than or equals 30	55 + 70 + 53 + 20 = 198
  More than or equals 40	70 + 53 + 20 = 143
  More than or equals 50	53 + 20 = 73
  More than or equals 60	20
  More than or equals 70	0

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

  Marks Below 20 Below 40 Below 60 Below 80 Below 100

  Number of students 17 22 29 37 50

  Form the frequency distribution table for the data.

  Solution:

  The frequency distribution table for given data.

  Marks	Number of students
  0 – 20	12
  20 – 40	22 – 17 = 5
  40 – 60	29 – 22 = 7
  60 – 80	37 – 29 = 8
  80 – 100	50 – 37 = 13

  15. Weekly income of 600 families is tabulated below :

  Weekly income Number of families

  (in Rs)

  0-1000 250

  1000-2000 190

  2000-3000 100

  3000-4000 40

  4000-5000 15

  5000-6000 5

  Total 600

  Compute the median income.

  Solution:

  Weekly Income	Number of families (fi)	Cumulative frequency (cf)
  0-1000	250	250
  1000-2000	190	250 + 190 = 400
  2000-3000	100	440 + 100 = 540
  3000-4000	40	540 + 40 = 580
  4000-5000	15	580 + 15 = 595
  5000-6000	5	595 + 5 = 600

  According to the question,

  n = 600

  ∴ n/2 = 600/2 = 300

  Cumulative frequency 440 lies in the interval 1000 – 2000.

  Hence, lower median class, l = 1000

  f = 190,

  c_f = 250,

  Class width, h = 1000

  And total observation n = 600

  

  = 1000 + 5000/19

  = 1000 + 263.15 = 1263.15

  Hence, the median income is Rs.1263.15.

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

  Speed (km/h) 85-100 100-115 115-130 130-145

  Number of players 11 9 8 5

  Calculate the median bowling speed.

  Solution:

  First we construct the cumulative frequency table

  Speed ( in km/h)	Number of players	Cumulative frequency
  85 – 100	11	11
  100 – 115	9	11 + 9 = 20
  115 – 130	8	20 + 8 = 28
  130 – 145	5	28 + 5 = 33

  It is given that, n = 33

  ∴ n/2 = 33/2 = 16.5

  Hence, the median class is 100 – 115.

  Where, lower limit(l) = 100

  Frequency (f) = 9

  Cumulative frequency (cf) = 11

  And class width(h) = 15

  

  = 100 + 82.5/9

  = 100 + 9.17

  = 109.17

  Hence, the median bowling speed is 109.17 km/h.

  17. The monthly income of 100 families are given as below :

  Income (in Rs) Number of families

  0-5000 8

  5000-10000 26

  10000-15000 41

  15000-20000 16

  20000-25000 3

  25000-30000 3

  30000-35000 2

  35000-40000 1

  Calculate the modal income.

  Solution:

  According to the data given,

  The highest frequency = 41,

  41 lies in the interval 10000 – 15000.

  Here, l = 10000, f_m = 41,f₁ = 26,f₂ = 16 and h = 5000

  

  = 10000 + 15×125

  = 10000 + 1875

  = 11875

  Hence, the modal income = Rs.11875 per month.

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

  Weight (in g) Number of packets

  200-201 12

  201-202 26

  202-203 20

  203-204 9

  204-205 2

  205-206 1

  Determine the modal weight.

  Solution:

  In the given data, the highest frequency is 26, which lies in the interval 201 – 202

  Here, l = 201,f_m = 26,f₁ = 12,f₂ = 20 and (class width) h = 1

  

  Hence, the modal weight = 201.7 g.

  19. Two dice are thrown at the same time. Find the probability of getting

  ASame number on both dice.
  Different numbers on both dice.

  Solution:

  Two dice are thrown at the same time.

  So, total number of possible outcomes = 36

  (i) Same number on both dice.

  Possible outcomes = (1,1), (2,2), (3, 3), (4, 4), (5, 5), (6, 6).

  Hence, number of possible outcomes = 6

  Therefore, the probability of getting same number on both dice = 6/36 = 1/6

  (ii) Different number on both dice.

  Hence, number of possible outcomes

  = 36 – Number of possible outcomes for same number on both dice

  = 36 – 6 = 30

  Therefore, the probability of getting different number on both dice = 30/36 = 5/6

  20. Two dice are thrown simultaneously. What is the probability that the sum of the numbers appearing on the dice is

  (i) 7? (ii) a prime number? (iii) 1?

  Solution:

  According to the question,

  Two dice are thrown simultaneously.

  So, that number of possible outcomes = 36

  (i) Sum of the numbers appearing on the dice is 7.

  So, the possible outcomes = (1, 6), (2,5), (3, 4), (4, 3), (5, 2), (6, 1).

  Hence, number of possible outcomes = 6

  ∴ the probability that the sum of the numbers appearing on the dice is 7 = 6/36 = 1/6

  (ii) Sum of the numbers appearing on the dice is a prime number i.e., 2, 3, 5, 7 and 11.

  So, the possible outcomes are (1, 1), (1,2), (2, 1), (1, 4), (2, 3), (3, 2), (4, 1), (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6) = (6, 5).

  Hence, number of possible outcomes = 15

  ∴ the probability that the sum of the numbers appearing on the dice is a prime number = 15/36 = 5/12

  (iii) Sum of the numbers appearing on the dice is 1.

  It is not possible, so its probability is zero.

  ∴ the probability that the sum of the numbers appearing on the dice is 1 = 0

  21. Two dice are thrown together. Find the probability that the product of the numbers on the top of the dice is

  (i) 6 (ii) 12 (iii) 7

  Solution:

  Number of total outcomes = 36

  (i) When product of the numbers on the top of the dice = 6.

  The possible outcomes = (1, 6), (2,3), (3, 2), (6, 1).

  Hence, number of possible ways = 4

  ∴ Probability that the product of the numbers on the top of the dice is 6= 4/36 = 1/9

  (ii) When product of the numbers on the top of the dice = 12.

  The possible ways are (2, 6), (3,4), (4, 3), (6, 2).

  Hence, number of possible ways = 4

  ∴ Probability that the product of the numbers on the top of the dice is 12= 4/36 = 1/9

  (iii) Product of the numbers on the top of the dice cannot be 7.

  Hence, the probability is zero.

  ∴ Probability that the product of the numbers on the top of the dice is 7 = 0

  ---

  To facilitate easy learning and aid students in understanding the chapter better, free NCERT Exemplar for Class 10 Maths Chapter 13 is provided. This can be downloaded in the form of a PDF. Students can also download other sources of learning from BYJU’S, such as notes, exemplar books, NCERT Maths Solutions for Class 10 and question papers, to prepare for their exams to score good marks. Solving sample papers and previous years’ question papers also gives an idea of question types asked in the board exam from Chapter 13 Statistics and Probability.

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  Frequently Asked Questions on NCERT Exemplar Solutions for Class 10 Maths Chapter 13

  Q1

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

  The topics and sub-topics covered in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths are listed below.
  1. Determining the mean of grouped data by a direct method, assumed mean method and step-deviation method
  2. Finding the mode of the given data
  3. To find the median of the grouped data
  4. Representation of cumulative frequency distribution graphically

  Q2

  Explain the concept of mean, as discussed in Chapter 13 of NCERT Exemplar Solutions for Class 10 Maths.

  Mean is nothing but the average of the given set of values. It denotes the equal distribution of values for a given data set. Central tendency is the statistical measure that recognises a single value as representative of the entire distribution. It provides an exact description of the whole data. It is the unique value that represents collected data. The mean, median and mode are the three commonly used measures of central tendency.

  Q3

  Why are NCERT Exemplar Solutions for Class 10 Maths Chapter 13 beneficial for students to score well in the board exam?

  NCERT Exemplar Solutions for Class 10 Maths Chapter 13 are provided by the subject experts at BYJU’S based on the understanding abilities of students. The solution module PDF can be downloaded from BYJU’S website based on the requirements of students. All important concepts are explained in simple language to help students ace the exam without fear. The solutions for all the problems present in the NCERT textbook help them to cross-check their answers and analyse their areas of weakness.

   

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