NCERT Solutions for Class 10 Maths Chapter 14 Statistics

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.

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

Answer :

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: $Midpoint(x_i)=\frac{upperlimit+lowerlimit}{2}$

The formula to find the mean is: ?? $\overline{x}=\frac{?f_i{x}_i}{?f_i}$

$=\frac{162}{20}=8.1$

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

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

Find the mean daily wages of the workers of the factory by using an appropriate method.

Answer :

The midpoint of the given interval is found by formula: $Midpoint(x_i)=\frac{upperlimit+lowerlimit}{2}$

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, $u_i=\frac{x_i?A}{h}$
$?u_i=\frac{x_i?150}{20}$

So, the formula to find out the mean is : $\overline{x}=A+h\frac{?f_i{x}_i}{?f_i}$
$=150+20\times \frac{?12}{50}$
$=150?4.8$
$=145.20$
$?$ 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.

Answer :

Given, $\overline{x}=18$

Thus, $\overline{x}=\frac{?f_i{x}_i}{?f_i}$

$?18=\frac{752+20f}{44+f}$

$?18(44+f)=(752+20f)$

$?792+18f=752+20f$

$?792?752=20f?18f$
$?40=2f$

$?f=20$

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.

Answer :

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

So, $u_i=\frac{x_i?A}{h}$
$?u_i=\frac{x_i?75.5}{h}$

$\overline{x}=A+h\frac{?f_i{x}_i}{?f_i}$

$=75.5+3\times \frac{4}{30}$
$=75.5+\frac{4}{10}$
$=75.9$

$?$ 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.

Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?

Answer :

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.

The formula to find out the Mean is:

$\overline{x}=A+h\frac{?f_i{d}_i}{?f_i}=57+3\times \frac{75}{400}$

$=57+0.1875=57.19$

$?$ 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.

Answer :

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

$\overline{x}=A+h\frac{?f_i{u}_i}{?f_i}$

$=225+50\times \frac{?7}{25}$

$=225?14=211$

$?$ 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:

Find the mean concentration of SO₂ in the air.

Answer :

Calculate mean by direct method:

The formula to find out the mean is

$\overline{x}=\frac{?f_i{x}_i}{?f_i}$

$=\frac{2.96}{30}$

$=0.099$ 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.

Answer :

The mean formula is,

$\overline{x}=\frac{?f_i{x}_i}{?f_i}$

$=\frac{499}{40}$

$=12.48$ 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.

Answer :

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

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

So, $u_i=\frac{x_i?A}{h}$

$=>u_i=\frac{x_i?70}{10}$

So,

$\overline{x}=A+h\frac{?f_i{x}_i}{?f_i}$

$=70+\frac{?2}{35}\times 10$

$=69.42$

$?$ The mean literacy part is 69.42

Q1 ) The following table shows the ages of the patients admitted in a hospital during a year:

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

Answer :

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

$Mode=l+(\frac{f_1?f_0}{2f_1?f_0?f_2})\times h$

$=35+\frac{23?21}{2\times 23?21?14}\times 10$

$=35+\frac{2}{46?21?14}\times 10$

$=35+\frac{2}{11}\times 10$

$=35+1.8=36.8$

Now, to calculate the Mean,

The formula for mean is :

$\overline{x}=\frac{?f_i{x}_i}{?f_i}$

$=\frac{2830}{80}=35.37$ 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:

Determine the modal lifetimes of the components.

Answer :

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

$Mode=l+(\frac{f_1?f_0}{2f_1?f_0?f_2})\times h$

$=60+(\frac{61?52}{2\times 61?52?38})\times 20$

$=60+(\frac{61?52}{122?52?38})\times 20$

$=60+\frac{9}{32}\times 20$

$=60+5.625=65.625$

$?$ 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 :

Answer :

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 $=l+(\frac{f_1?f_0}{2f_1?f_0?f_2})\times h$

$=1500+(\frac{40?24}{2\times 40?24?33})\times 500$

$=1500+\frac{40?24}{80?24?33}\times 500$

$=1500+\frac{16}{23}\times 500$

$=1500+347.83=1847.83$

$?$ Modal monthly expenditure = Rs. 1847.83

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

The formula to calculate the mean,

$\overline{x}=A+h\frac{?f_i{x}_i}{?f_i}$

$=2750+(\frac{?35}{200})\times 500$

$=2750?87.50=2662.50$

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

Answer :

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 $=l+(\frac{f_1?f_0}{2f_1?f_0?f_2})\times h$

$=30+(\frac{10?9}{2\times 10?9?3})\times 5$

$=30+\frac{10?9}{20?9?3}\times 5$

$=30+\frac{1}{8}\times 5=30+0.625$

$=30.625=30.6$

Calculation of mean :

$\overline{x}=\frac{?f_i{x}_i}{?f_i}$

$=\frac{1022.5}{35}=29.2$

$?$ 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.

Find the mode of the data.

Answer :

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 $=l+(\frac{f_1?f_0}{2f_1?f_0?f_2})\times h$

$=4000+(\frac{18?4}{2\times 18?4?9})\times 1000$

$=4000+\frac{18?4}{36?4?9}\times 1000$

$=4000+\frac{14}{23}\times 1000$

$=4000+608.695=4608.7$

$?$ 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:

Answer :

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 $=l+(\frac{f_1?f_0}{2f_1?f_0?f_2})\times h$

$=40+(\frac{20?12}{40?12?11})\times 10$

$=40+\frac{8}{17}\times 10$

$=40+4.7=44.7$

? Mode = 44.7 cars

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.

Answer :

We have n = 68
$?\frac{n}{2}=\frac{68}{2}=34$

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 $\frac{n}{2}=34,$
l = 125 ,
c.f. = 22 ,
f = 20 , h = 20.

Median $=l+(\frac{\frac{n}{2}?cf}{f})\times h$,

Median $=125+(\frac{34?22}{20})\times 20$
$=125+12=137$ units

To calculate the mode :

Modal class = $125?145,$
$f_1=20,$
$f_0=13,$
$f_2=14,$ and
$h=20$

Mode formula :

Mode = $l+\frac{(f_1?f_0)}{(2f_1?f_0?f_2)}\times h$

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

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

$=125+10.77=135.77$

Therefore, mode = 135.77

Calculation of mean :

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

so $u_i=\frac{x_i?A}{h}=\frac{x_i?135}{20}$

$\overline{x}=A+h\frac{?f_i{x}_i}{?f_i}$

$=135+20(\frac{7}{68})$

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.

Answer :

Given that median is 28.5 and $n=?f_i=60$

Here n = 60
$?\frac{n}{2}=30$

Since the median is given to be 28.5, thus the median class is 20-30.

$?$ l = 20, h = 10, f = 20 and cf = 5 + x

$?$ median $=l+(\frac{\frac{n}{2}?cf}{f})\times h$

$?28.5=20+\frac{30?5?x}{20}\times 10$