# Statistics : Exercise - 14.1 (Mathematics NCERT Class 10th)

Q.1     A survey was conducted by a group of students as a part of their environment awareness
programmer, 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 ?
Sol.      Calculation of Mean Hence, mean $\bar x = {1 \over n}\Sigma {f_i}{x_i}$
= ${1 \over {20}} \times 162 = 8.1$ plants
We have used direct method because numerical values of ${x_i}$ and ${f_i}$ are small.

Q.2     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.
Sol.      Let the assumed mean, A =150; class-interval, h = 20
so         ${u_i} = {{{x_i}-A} \over h} = {{{x_i}-150} \over {20}}$
We construct the table : $\bar x = A + h{{\Sigma {f_i}{u_i}} \over {\Sigma {f_i}}} = 150 + 20 \times {{ - 12} \over {50}}$
= 150 – 4.8 = 145.2
Hence, mean = Rs 145.20 .

Q.3     The following distribution shows the daily pocket allowance of children locality mean. The mean pocket allowance is Rs 18. Find the missing frequency f. Sol.       Let the assumed mean, A = 16, class interval h = 2
So       ${u_i} = {{{x_i}-A} \over h} = {{{x_i}-16} \over 2}$ We have, ${\bar x}$ = 18, A = 16 and h = 2
Therefore, $\bar x = A + h\left( {{1 \over N}\Sigma {f_i}{u_i}} \right)$
$\Rightarrow 18 = 16 + 2\left( {{{2f + 24} \over {f + 44}}} \right)$
$\Rightarrow 2 = 2\left( {{{2f + 24} \over {f + 44}}} \right)$
$\Rightarrow f + 44 = 2f + 24$
$\Rightarrow$2ff = 44 – 24
$\Rightarrow f = 20$
Hence, the missing frequency is 20.

Q.4      Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarised as follows. Find the mean heart beats per minute for these women, choosing a suitable method. .
Sol.       Let the assumed mean, A = 75.5, class interval
h = 3, so     ${u_i} = {{{x_i}-A} \over h} = {{{x_i}-75.5} \over 3}$ $\bar x = A + h{{\Sigma {f_i}{u_i}} \over {\Sigma {f_i}}} = 75.5 + 3 \times {4 \over {30}}$
= 75.5 + 0.4 = 75.9
Hence, the mean heart beats per minute for these women is 75.9.

Q.5     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 ?

Sol.      Here, the class intervals are formed by the exclusive method. If we make the series an inclusive one the mid-values remain  same. So there is no need to convert the series.
Let the assumed mean be A = 60 and h = 3
so      ${u_i} = {{{x_i}-A} \over h} = {{{x_i}-60} \over 3}$
Calculation of Mean Therefore $\bar x = A + h{{\Sigma {f_i}{u_i}} \over {\Sigma {f_i}}} = 60 + 3 \times {{ - 375} \over {400}}$
= 60 – 2.8125 = 57.1875 = 57.19(nearly)
Hence, average number of mangoes per box = 57.19 (nearly)

Q.6      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.
Sol.       Let the assumed mean, A = 225, class interval, h = 50
so      ${u_i} = {{{x_i} - A} \over h} = {{{x_i} - 225} \over {50}}$
We construct the following table : $\bar x = A + h{{\Sigma {f_i}{u_i}} \over {\Sigma {f_i}}}$
= 225 +50$\times {{ - 7} \over {25}}$
= 225 – 14 = 211
Hence, the mean daily expenditure of food is Rs 211.

Q.7     To find the concentration of $S{O_2}$ 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 $S{O_2}$ in the air.
Sol.       Calculation of mean by direct method Hence, $\bar x = {{\sum {f_i}{x_i}} \over {\sum {f_i}}}$
= ${{2.96} \over {30}}$ = 0.099 ppm

Q.8     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. Sol.      Here, the class size varies, and ${{x_i}}$ 's are small. Let us apply the direct method here. Therefore, mean, $\bar x = {{\Sigma {f_i}{x_i}} \over {\Sigma {f_i}}}$
= ${{499} \over {40}}$ = 12.475
Hence,      mean = 12.48 days

Q.9     The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate. Sol.
Let the assumed mean be A = 70 and h = 10

so      ${u_i} = {{{x_i} - 70} \over {10}}$ Therefore,  $\bar x = A + h\, \times {{\Sigma {f_i}{u_i}} \over {\Sigma {f_i}}}$
= 70 + 10 $\times {{ - 2} \over {35}}$
= 70 – 0.57 = 69.43
Thus, mean literacy rate is 69.43% .