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

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

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Exercise 14.1 Solutions 9 Question (9 long)

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Exercise 14.4 Solutions 3 Question (3 long)

### Access Answers to NCERT Class 10 Maths Chapter 14 – Statistics Exercise 14.2

1. The following table shows the ages of the patients admitted in 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 – 45,

l = 35,

class width (h) = 10,

fm = 23,

f1 = 21 and f2 = 14

The formula to find the mode is

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

Mode = 35+(20/11) = 35+1.8

Mode = 36.8 year

So the mode of the given data = 36.8 year

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 = x̄ = ∑fixi /∑fi

= 2830/80

= 35.37 years

Therefore, the mean of the given data = 35.37 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 60–80.

l = 60,

The frequencies are:

fm = 61, f1 = 52, f2 = 38 and h = 20

The formula to find the mode is

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the formula, we get

Mode =60+[(61-52)/(122-52-38)]×20

Mode = 60+((9 x 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 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-f1)/(2fm-f1-f2)]×h

Substitute the values in the formula, we get

Mode =1500+[(40-24)/(80-24-33)]×500

Mode = 1500+((16×500)/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 – 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 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 = x̄ = a +(∑fiui /∑fi)×h

Substitute the values in the given formula

= 2750+(-35/200)×500

= 2750-87.50

= 2662.50

So, the mean monthly expenditure of the families = Rupees 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 – 35,

l = 30,

Class width (h) = 5,

fm = 10, f1 = 9 and f2 = 3

Mode Formula:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values in the given formula

Mode = 30+((10-9)/(20-9-3))×5

Mode = 30+(5/8) = 30+0.625

Mode = 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 52.5 105.5 Sum fi = 35 Sum fixi = 1022.5

Mean = x̄ = ∑fixi /∑fi

= 1022.5/35

= 29.2

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 – 5000,

l = 4000,

class width (h) = 1000,

fm = 18, f1 = 4 and f2 = 9

Mode Formula:

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values

Mode = 4000+((18-4)/(36-4-9))×1000

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 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 – 50, l = 40,

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

Mode = l+ [(fm-f1)/(2fm-f1-f2)]×h

Substitute the values

Mode = 40+((20-12)/(40-12-11))×10

Mode = 40 + (80/17) = 40 + 4.7 = 44.7

Thus, the mode of the given data is 44.7 cars

In this exercise, students will solve questions based on finding the mode of grouped data along with finding 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.