NCERT Solutions for Exercise 14.2 Class 10 Maths Chapter 14 - Statistics

Q2 The following data gives information on the observed lifetimes (in hours) of 225 electrical components :

Determine the modal lifetimes of the components.

Answer:

To find the modal class: The class having the maximum frequency is the modal class. And, the maximum frequency is 61, and hence the modal class = 60-80

Thus, Lower limit (l) of modal class = 60, class size (h) = 20

Frequency ( $f_1$ ) of the modal class = 61

Frequency ( $f_0$ ) of class preceding the modal class = 52

Frequency ( $f_2$ ) of class succeeding the modal class = 38.

Therefore, $Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =60+\left(\frac{61-52}{2(61)-52-38}\right).20 \\ =60+\frac{9}{32}.20 \end{gathered}$$

$=65.62$

Thus, the modal lifetime of 225 electrical components is 65.62 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:

To find the modal class: The class having the maximum frequency is the modal class. And, the maximum frequency is 40, and hence the modal class = 1500-2000

Thus, Lower limit (l) of modal class = 1500, class size (h) = 500

Frequency ( $f_1$ ) of the modal class = 40

Frequency ( $f_0$ ) of class preceding the modal class = 24

Frequency ( $f_2$ ) of class succeeding the modal class = 33.

Therefore, $Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =1500+\left(\frac{40-24}{2(40)-24-33}\right).500 \\ =1500+\frac{16}{23}.500 \end{gathered}$$

$=1847.82$

Thus, the Mode of the data is Rs. 1847.82

Now,

Let the assumed mean be a = 2750 and h = 500

Mean,

$\bar{x}=a+\frac{\sum f_i{u}_i}{\sum f_i}\times h$

$=2750+\frac{-35}{200}\times 500=2750-87.5=2662.50$

Thus, the Mean monthly expenditure 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:

To find the modal class: The class having the maximum frequency is the modal class. And, the maximum frequency is 10, and hence the modal class = 30-35

Lower limit (l) of modal class = 30, class size (h) = 5

Frequency ( $f_1$ ) of the modal class = 10

Frequency ( $f_0$ ) of class preceding the modal class = 9

Frequency ( $f_2$ ) of class succeeding the modal class = 3

Therefore, $Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =30+\left(\frac{10-9}{2(10)-9-3}\right).5 \\ =30+\frac{1}{8}.5 \end{gathered}$$

$=30.625$

Thus, the Mode of the data is 30.625

Now,

Let the assumed mean be a = 32.5 and h = 5

Mean,

$\bar{x}=a+\frac{\sum f_i{u}_i}{\sum f_i}\times h$

$=32.5+\frac{-23}{35}\times 5=29.22$

Thus, the Mean of the data is 29.22

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:

To find the modal class: The class having the maximum frequency is the modal class. And, the maximum frequency is 18, and hence the modal class = 4000-5000

Lower limit (l) of modal class = 4000, class size (h) = 1000

Frequency ( $f_1$ ) of the modal class = 18

Frequency ( $f_0$ ) of class preceding the modal class = 4

Frequency ( $f_2$ ) of class succeeding the modal class = 9

Therefore, $Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =4000+\left(\frac{18-4}{2(18)-4-9}\right).1000 \\ =4000+\frac{14}{23}.1000 \end{gathered}$$

$=4608.70$

Thus, the Mode of the data is 4608.70

Q6 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:

To find the modal class: The class having the maximum frequency is the modal class. And, the maximum frequency is 20, and hence the modal class = 40-50

Lower limit (l) of modal class = 40, class size (h) = 10

Frequency ( $f_1$ ) of the modal class = 20

Frequency ( $f_0$ ) of class preceding the modal class = 12

Frequency ( $f_2$ ) of class succeeding the modal class = 11

Therefore, $Mode=l+\left(\frac{f_1-f_0}{2f_1-f_0-f_2}\right).h$

$$\begin{gathered} =40+\left(\frac{20-12}{2(20)-12-11}\right).10 \\ =40+\frac{8}{17}.10 \end{gathered}$$

$=44.70$

Thus, the Mode of the data is 44.70