NCERT Solutions for Class 10 Maths Chapter 14 Statistics  Written by Team Trustudies
Updated at 2021-05-07

NCERT solutions for class 10 Maths Chapter 14 Statistics Exercise - 14.1

Q1 ) A survey was conducted by a group of students as a part of their environment awareness program, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.

 Number of Plants 0-2 2-4 4-6 6-8 8-10 10-12 12-14 Number of Houses 1 2 1 5 6 2 3

Which method did you use for finding the mean, and why?

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

In order to find the mean value, we will use direct method because the numerical value of ${f}_{i}$ and ${x}_{i}$ are small.

The midpoint of the given interval is found by formula:

 No. of plants (Class interval) No. of houses (Frequency $\left({f}_{i}\right)$ ) Mid-point $\left({x}_{i}\right)$ ${f}_{i}{x}_{i}$ 0-2 1 1 1 2-4 2 3 6 4-6 1 5 5 6-8 5 7 35 8-10 6 9 54 10-12 2 11 22 12-14 3 13 39

The formula to find the mean is: ??

$?$ The mean number of plants per house is 8.1

Q2 ) Consider the following distribution of daily wages of 50 workers of a factory.

 Daily wages (in Rs.) 100-120 120-140 140-160 160-180 180-200 Number of workers 12 14 8 6 10
Find the mean daily wages of the workers of the factory by using an appropriate method.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

The midpoint of the given interval is found by formula:

In this case, the value of mid-point ( ${x}_{i}$) is very large,so let us assume the mean value, A = 150 and class interval is h = 20.

So,

 Daily wages (Class interval) Number of workers (frequency (${f}_{i}$) ) Mid-point ( ${x}_{i}$ ) ${f}_{i}{u}_{i}$ 100-120 12 110 -2 -24 120-140 14 130 -1 -14 140-160 8 150 0 0 160-180 6 170 1 6 180-200 10 190 2 20 Total

So, the formula to find out the mean is :

$?$ Mean daily wage of the workers = Rs. 145.20

Q3 ) The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.

 Daily Pocket Allowence(in Rs) 11-13 13-15 15-17 17-19 19-21 21-23 23-35 Number of children 7 6 9 13 f 5 4

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Given,

 Class interval Number of children (frequency (${f}_{i}$) ) Mid-point $\left({x}_{i}\right)$ ${f}_{i}{u}_{i}$ 11-13 7 12 84 13-15 6 14 84 15-17 9 16 144 17-19 13 18 234 19-21 f 20 20f 21-23 5 22 110 23-25 4 24 96 Total

Thus,

So, the missing frequency, f = 20.

Q4 ) Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.

 Number of heartbeats per minute 65-68 68-71 71-74 74-77 77-80 80-83 83-86 Number of women 2 4 3 8 7 4 2

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

From the given data, let us assume the mean as A = 75.5
Class size (h) = 3

So,

 Class Interval Number of women ( ${f}_{i}$ ) Mid-point ( ${x}_{i}$ ) ${f}_{i}{u}_{i}$ 65-68 2 66.5 -3 -6 68-71 4 69.5 -2 -8 71-74 3 72.5 -1 -3 74-77 8 75.5 0 0 77-80 7 78.5 1 7 80-83 4 81.5 3 8 83-86 2 84.5 3 6

$?$ The mean heart beats per minute for these women is 75.9

Q5 ) In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.

 Number of mangoes 50-52 53-55 56-58 59-61 62-64 Number of boxes 15 110 135 115 25
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.5 from the lower limit as the gap between two intervals are 1

Here, assumed mean (A) = 57, Class size (h)
= 3

Here, the step deviation is used because the frequency values are big.

 Class Interval Number of boxes $\left({f}_{i}\right)$ Mid-point $\left({x}_{i}\right)$ ${f}_{i}{d}_{i}$ 49.5-52.5 15 51 -6 90 52.5-55.5 110 54 -3 -330 55.5-58.5 135 57 0 0 58.5-61.5 115 60 3 345 61.5-64.5 25 63 6 150

The formula to find out the Mean is:

$?$ The mean number of mangoes kept in a packing box is 57.19

Q6 ) The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.

 Daily expenditure(in Rs) 100-150 150-200 200-250 250-300 300-350 Number of households 4 5 12 2 2

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Let us assume the mean (A) = 225, Class size (h) = 50

 Class Interval Number of households $\left({f}_{i}\right)$ Mid-point $\left({x}_{i}\right)$ ${f}_{i}{u}_{i}$ 100-150 4 125 -100 -2 -8 150-200 5 175 -50 -1 -5 200-250 12 225 0 0 0 250-300 2 275 50 1 2 300-350 2 325 100 2 4

$?$ The mean daily expenditure on food is 211

Q7 ) To find out the concentration of SO2 in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:

 Concentration of SO2 ( in ppm) Frequency 0.00 – 0.04 4 0.04 – 0.08 9 0.08 – 0.12 9 0.12 – 0.16 2 0.16 – 0.20 4 0.20 – 0.24 2
Find the mean concentration of SO2 in the air.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

Calculate mean by direct method:

 Concentration of SO2 (in ppm) Frequency $\left({f}_{i}\right)$ Mid-point $\left({x}_{i}\right)$ ${f}_{i}{x}_{i}$ 0.00 - 0.04 4 0.02 0.08 0.04 - 0.08 9 0.06 0.54 0.08 - 0.12 9 0.10 0.90 0.12 - 0.16 2 0.14 0.28 0.16 - 0.20 4 0.18 0.72 0.20 - 0.24 2 0.20 0.40 Total

The formula to find out the mean is

ppm

$?$ The mean concentration of SO2 in air is 0.099 ppm.

Q8 ) A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.

 Number of days 0-6 6-10 10-14 14-20 20-28 28-38 38-40 Number of students 11 10 7 4 4 3 1

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

 Class interval Frequency $\left({f}_{i}\right)$ Mid-point $\left({x}_{i}\right)$ ${f}_{i}{x}_{i}$ 0-6 11 3 33 6-10 10 8 80 10-14 7 12 84 14-20 4 17 68 20-28 4 24 96 28-38 3 33 99 38-40 1 39 39

The mean formula is,

days

$?$ The mean number of days a student was absent = 12.48.

Q9 ) The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.

 Literacy rate (in %) 45-55 55-65 65-75 75-85 85-98 Number of cities 3 10 11 8 3

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

In this case, the value of mid-point (xi) is very large,

so let us assume the mean value, A = 70 and class interval is h = 10.

So,

 Class Interval Frequency $\left({f}_{i}\right)$ $\left({x}_{i}\right)$ ${f}_{i}{u}_{i}$ 45-55 3 50 -20 -2 -6 55-65 10 60 -10 -1 -10 65-75 11 70 0 0 0 75-85 8 80 10 1 8 85-95 3 90 20 2 6

So,

$?$ The mean literacy part is 69.42

NCERT solutions for class 10 Maths Chapter 14 Statistics Exercise - 14.2

Q1 ) 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.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

The class 35-45 has the maximum frequency, therefore, this is the modal class.

Here l = 35 , h = 10 , f1 = 23 , f0 = 21 , f2 =14

Now, let us substitute these values in the formula

Now, to calculate the Mean,

 Class Interval Frequency ( ${f}_{i}$ ) Mid-point ( ${x}_{i}$ ) ${f}_{i}{x}_{i}$ 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

The formula for mean is :

years

$?$ The mean of the given data = 35.37 years

Q2 ) 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.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

The class 60-80 has the maximum frequency, therefore, this is the modal class.

Here l = 60 , h = 20 , f1 = 61 , f0 = 52 , and f2 = 38

Now, let us substitute these values in the formula

$?$ The modal lifetimes of the components is 65.625 hours.

Q3 ) 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

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

The class 1500 – 2000 has the maximum frequency. Therefore, it is the modal class.

Here l = 1500,
h = 500,
f1 =40 ,
f0 =24 and
f2 =33

Now, let us substitute these values in the formula

Mode

$?$ Modal monthly expenditure = Rs. 1847.83

Let the assumed mean be A = 2750 and h = 500.

 Class Interval ${f}_{i}$ ${x}_{i}$ ${f}_{i}{u}_{i}$ 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

The formula to calculate the mean,

So, the mean monthly expenditure of the families is Rs 2662.50

Q4 ) 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

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

The class 30-35 has the maximum frequency, therefore, this is the modal class.

Here l = 30,
h = 5, f1 = 10,
f0 = 9 and
f2 = 3

Now, let us substitute these values in the formula

Mode

Calculation of mean :

 Class Interval Frequency $\left({f}_{i}\right)$ Mid-point $\left({x}_{i}\right)$ ${f}_{i}{x}_{i}$ 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

$?$ mean = 29.2

Q5 ) 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.

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

The class 4000-5000 has the maximum frequency ,
therefore, this is the modal class.

Here l = 4000,
h = 1000,
f1 = 18,
f0 = 4 and
f2 = 9

Now, let us substitute these values in the formula

Mode

$?$ 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 summarised 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

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

The class 40-50 has the maximum frequency, therefore, this is the modal class.

Here l = 40, h =10, f1 = 20, f0 = 12 and f2 = 11

Now, let us substitute these values in the formula
Mode

? Mode = 44.7 cars

NCERT solutions for class 10 Maths Chapter 14 Statistics Exercise - 14.3

Q1 ) The following frequency distribution gives the monthly consumption of an electricity of 68 consumers in a locality. Find the median, mean and mode of the data and compare them.

 Monthly consumption(in units) No. of customers 65-85 4 85-105 5 105-125 13 125-145 20 145-165 14 165-185 8 185-205 4

NCERT Solutions for Class 10 Maths Chapter 14 Statistics

 Class Interval Frequency Cumulative frequency 65-85 4 4 85-105 5 9 105-125 13 22 125-145 20 42 145-165 14 56 165-185 8 64 185-205 4 68 N=68

We have n = 68

The cumulative frequency just greater than $\frac{n}{2}$ , is 42 and the corresponding class is 125-145.

Thus, 125-145 is the median class such that
l = 125 ,
c.f. = 22 ,
f = 20 , h = 20.

Median ,

Median
units

To calculate the mode :

Modal class = $125?145,$
${f}_{1}=20,$
${f}_{0}=13,$
${f}_{2}=14,$ and

Mode formula :

Mode = $l+\frac{\left({f}_{1}?{f}_{0}\right)}{\left(2{f}_{1}?{f}_{0}?{f}_{2}\right)}×h$

Mode = $125+\frac{\left(20?13\right)}{\left(40?13?14\right)}×20$

$=125+\frac{140}{13}$

$=125+10.77=135.77$

Therefore, mode = 135.77

Calculation of mean :

Let the assumed mean, A = 135, class interval, h = 20

so

 Class Interval ${f}_{i}$ ${x}_{i}$ ${f}_{i}{u}_{i}$ 65-85 4 75 -60 -3 -12 85-105 5 95 -40 -2 -10 105-125 13 115 -20 -1 -13 125-145 20 135 0 0 0 145-165 14 155 20 1 14 165-185 8 175 40 2 16 185-205 4 195 60 3 12

Mean = 137.05

In this case, mean, median and mode are more/less equal in this distribution.

Q2 ) If the median of a distribution given below is 28.5 then, find the value of an x &y.

 Class Interval Frequency 0-10 5 10-20 x 20-30 20 30-40 15 40-50 y 50-60 5 Total 60

NCERT Solutions for Class 10 Maths Chapter 14 Statistics