# NCERT Solutions for class 10 Maths Chapter 14 - Statistics Exercise 14.2

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.

NCERT Solutions for Class 10 Maths Chapter 14 â€“ Statistics Exercise 14.2 have been provided here to help students prepare for their CBSE Exam. The problems in NCERT Solutions have been answered by our subject experts, covering the NCERT syllabus and guidelines prescribed by the Central Board of Secondary Education.

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## NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.2

### Access Other Exercise Solutions of Class 10 Maths Chapter 14 â€“ Statistics

Exercise 14.1 Solutions 9 Question (9 long)

Exercise 14.3 Solutions 7 Question (7 long)

Exercise 14.4 Solutions 3 Question (3 long)

### Access Answers to NCERT Class 10 Maths Chapter 14 â€“ Statistics Exercise 14.2

1. The following table shows the ages of the patients admitted to a hospital during a year:

 Age (in years) 5-15 15-25 25-35 35-45 45-55 55-65 Number of patients 6 11 21 23 14 5

Find the mode and the mean of the data given above. Compare and interpret the two

measures of central tendency.

Solution:

To find out the modal class, let us the consider the class interval with high frequency.

Here, the greatest frequency = 23, so the modal class = 35 â€“ 45,

Lower limit of modal class = l = 35,

class width (h) = 10,

fm = 23,

f1 = 21 and f2 = 14

The formula to find the mode is

Mode = l + [(fm â€“ f1)/ (2fm â€“ f1 â€“ f2)] Ã— h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]Ã—10

= 35 + (20/11)

= 35 + 1.8

= 36.8 years

So the mode of the given data = 36.8 years

Calculation of Mean:

First find the midpoint using the formula, xi = (upper limit +lower limit)/2

 Class Interval Frequency (fi) Mid-point (xi) fixi 5-15 6 10 60 15-25 11 20 220 25-35 21 30 630 35-45 23 40 920 45-55 14 50 700 55-65 5 60 300 Sum fi = 80 Sum fixi = 2830

The mean formula is

Mean = xÌ„ = âˆ‘fixi /âˆ‘fi

= 2830/80

= 35.375 years

Therefore, the mean of the given data = 35.375 years

2. The following data gives the information on the observed lifetimes (in hours) of 225

electrical components:

 Lifetime (in hours) 0-20 20-40 40-60 60-80 80-100 100-120 Frequency 10 35 52 61 38 29

Determine the modal lifetimes of the components.

Solution:

From the given data the modal class is 60â€“80.

Lower limit of modal class = l = 60,

The frequencies are:

fm = 61, f1 = 52, f2 = 38 and h = 20

The formula to find the mode is

Mode = l+ [(fm â€“ f1)/(2fm â€“ f1 â€“ f2)] Ã— h

Substitute the values in the formula, we get

Mode = 60 + [(61 â€“ 52)/ (122 â€“ 52 â€“ 38)] Ã— 20

Mode = 60 + [(9 Ã— 20)/32]

Mode = 60 + (45/8) = 60 + 5.625

Therefore, modal lifetime of the components = 65.625 hours.

3. The following data gives the distribution of total monthly household expenditure of 200

families of a village. Find the modal monthly expenditure of the families. Also, find the

mean monthly expenditure:

 Expenditure (in Rs.) Number of families 1000-1500 24 1500-2000 40 2000-2500 33 2500-3000 28 3000-3500 30 3500-4000 22 4000-4500 16 4500-5000 7

Solution:

Given data:

Modal class = 1500-2000,

l = 1500,

Frequencies:

fm = 40 f1 = 24, f2 = 33 and

h = 500

Mode formula:

Mode = l + [(fm â€“ f1)/ (2fm â€“ f1 â€“ f2)] Ã— h

Substitute the values in the formula, we get

Mode = 1500 + [(40 â€“ 24)/ (80 â€“ 24 â€“ 33)] Ã— 500

Mode = 1500 + [(16 Ã— 500)/23]

Mode = 1500 + (8000/23) = 1500 + 347.83

Therefore, modal monthly expenditure of the families = Rupees 1847.83

Calculation for mean:

First find the midpoint using the formula, xi =(upper limit +lower limit)/2

Let us assume a mean, (a) be 2750.

 Class Interval fi xi di = xi â€“ a ui = di/h fiui 1000-1500 24 1250 -1500 -3 -72 1500-2000 40 1750 -1000 -2 -80 2000-2500 33 2250 -500 -1 -33 2500-3000 28 2750 = a 0 0 0 3000-3500 30 3250 500 1 30 3500-4000 22 3750 1000 2 44 4000-4500 16 4250 1500 3 48 4500-5000 7 4750 2000 4 28 fi = 200 fiui = -35

The formula to calculate the mean,

Mean = xÌ„ = a +(âˆ‘fiui /âˆ‘fi) Ã— h

Substitute the values in the given formula

= 2750 + (-35/200) Ã— 500

= 2750 â€“ 87.50

= 2662.50

So, the mean monthly expenditure of the families = Rs. 2662.50

4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures

 No of students per teacher Number of states / U.T 15-20 3 20-25 8 25-30 9 30-35 10 35-40 3 40-45 0 45-50 0 50-55 2

Solution:

Given data:

Modal class = 30 â€“ 35,

l = 30,

Class width (h) = 5,

fm = 10, f1 = 9 and f2 = 3

Mode Formula:

Mode = l + [(fm â€“ f1)/ (2fm â€“ f1 â€“ f2)] Ã— h

Substitute the values in the given formula

Mode = 30 + [(10 â€“ 9)/ (20 â€“ 9 â€“ 3)] Ã— 5

= 30 + (5/8)

= 30 + 0.625

= 30.625

Therefore, the mode of the given data = 30.625

Calculation of mean:

Find the midpoint using the formula, xi =(upper limit +lower limit)/2

 Class Interval Frequency (fi) Mid-point (xi) fixi 15-20 3 17.5 52.5 20-25 8 22.5 180.0 25-30 9 27.5 247.5 30-35 10 32.5 325.0 35-40 3 37.5 112.5 40-45 0 42.5 0 45-50 0 47.5 0 50-55 2 52.5 105.0 Sum fi = 35 Sum fixi = 1022.5

Mean = xÌ„ = âˆ‘fixi /âˆ‘fi

= 1022.5/35

= 29.2 (approx)

Therefore, mean = 29.2

5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.

 Run Scored Number of Batsman 3000-4000 4 4000-5000 18 5000-6000 9 6000-7000 7 7000-8000 6 8000-9000 3 9000-10000 1 10000-11000 1

Find the mode of the data.

Solution:

Given data:

Modal class = 4000 â€“ 5000,

l = 4000,

class width (h) = 1000,

fm = 18, f1 = 4 and f2 = 9

Mode Formula:

Mode = l + [(fm â€“ f1)/ (2fm â€“ f1 â€“ f2)] Ã— h

Substitute the values

Mode = 4000 + [(18 â€“ 4)/ (36 â€“ 4 â€“ 9)] Ã— 1000

= 4000 + (14000/23)

= 4000 + 608.695

= 4608.695

= 4608.7 (approximately)

Thus, the mode of the given data is 4608.7 runs.

6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:

 Number of cars Frequency 0-10 7 10-20 14 20-30 13 30-40 12 40-50 20 50-60 11 60-70 15 70-80 8

Solution:

Given Data:

Modal class = 40 â€“ 50, l = 40,

Class width (h) = 10, fm = 20, f1 = 12 and f2 = 11

Mode = l + [(fm â€“ f1)/(2fm â€“ f1 â€“ f2)] Ã— h

Substitute the values

Mode = 40 + [(20 â€“ 12)/ (40 â€“ 12 â€“ 11)] Ã— 10

= 40 + (80/17)

= 40 + 4.7

= 44.7

Thus, the mode of the given data is 44.7 cars.

In this exercise of NCERT Class 10 Solutions, students will solve questions based on finding the mode of grouped data and the mean value. In a grouped frequency distribution, we cannot find the mode with the help of given frequencies, but we can locate a class with the maximum frequency called the modal class. The mode is a value inside the modal class, and its formula is,

where l = lower limit of the modal class,

h = size of the class interval (assuming all class sizes to be equal),

f1 = frequency of the modal class,

f0= frequency of the class preceding the modal class,

f2 = frequency of the class succeeding the modal class.

Approach us to practise the exercise problems with their solutions for class 10 Maths, Chapter 14. Also, get some more materials, such as notes, books, questions papers, etc., to make practice more effective.