KC Sinha Mathematics Solution Class 10 Chapter 6 Statistics Exercise 6.2

Exercise 6.1
Exercise 6.2
Exercise 6.3
Exercise 6.4

Exercise 6.2

Question 1

Find mode of the following data:

64, 61, 62, 62, 63, 61, 63, 64, 64, 60, 65, 63, 64, 65, 66, 64
Sol :

Here, we can see that 64 observation has the maximum frequency.
Hence, Mode = 64

Question 2

Find the mode of the following distribution:

Sol :
Here, the marks 8 has the maximum frequency ‘16’.
∴ Mode = 8

Question 3

Find the mode of the following data:

Sol :
Here, the class interval 12 has the maximum frequency ‘9’.
∴ Mode = 12

Question 4

A survey conducted on 20 households in a locality by a group of students resulted in the following frequency table for the number of family members in a household.

Find the mode of this data.
Sol :
Here, the maximum class frequency is 8, and the class corresponding to this frequency is 3 – 5.
So, the modal class is 3 – 5.
Now, modal class = 3 – 5, lower limit (l) of modal class = 3, class size(h) = 2
frequency (f₁) of the modal class = 8
frequency (f₀) of class preceding the modal class = 7
frequency (f₂) of class succeeding the modal class = 2
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=3+\left(\frac{8-7}{2 \times 8-7-2}\right) \times 2$
$=3+\frac{2}{7} \times 2$
= 3.286

Question 5

Find the mode of the following distribution:

Sol :
Here, the maximum number of students i.e. 71 have got marks in the interval 60 – 70.
So, the modal class is 60 – 70.
Now, modal class = 60 – 70, lower limit (l) of modal class = 60, class size(h) = 10
frequency (f₁) of the modal class = 71
frequency (f₀) of class preceding the modal class = 31
frequency (f₂) of class succeeding the modal class = 52
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=60+\left(\frac{71-31}{2 \times 71-31-52}\right) \times 10$
$=60+\frac{40}{59} \times 10$
= 60 + 6.78
= 66.78

Question 6

Find the mode of the following distribution:

Sol :
Here, the maximum class frequency is 28, and the class corresponding to this frequency is 30 – 40.
So, the modal class is 30 – 40.
Now, modal class = 30 – 40, lower limit (l) of modal class = 30, class size(h) = 10
frequency (f₁) of the modal class = 28
frequency (f₀) of class preceding the modal class = 25
frequency (f₂) of class succeeding the modal class = 25
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=30+\left(\frac{28-25}{2 \times 28-25-25}\right) \times 10$
$=30+\frac{3}{6} \times 10$
= 30 + 5
= 35

Question 7

The given distribution shows the number of runs scored by some top batsman of the world in one-day international cricket matches.

Find the mode of the data.
Sol :
Here, the maximum number of batsman i.e. 18 have scored the runs in the interval 4000 – 5000.
So, the modal class is 4000 – 5000 .
Now, modal class = 4000 – 5000,
lower limit (l) of modal class = 4000,
class size(h) = 1000
frequency (f₁) of the modal class = 18
frequency (f₀) of class preceding the modal class = 4
frequency (f₂) of class succeeding the modal class = 9
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=4000+\left(\frac{18-4}{2 \times 18-4-9}\right) \times 1000$
$=4000+\frac{14}{36-13} \times 1000$
= 4000 + 608.7

= 4608.7

Question 8

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:

Sol :
Here, the maximum class frequency is 20, and the class corresponding to this frequency is 40 – 50.
So, the modal class is 40 – 50.
Now, modal class = 40 – 50, lower limit (l) of modal class = 40, class size(h) = 10
frequency (f₁) of the modal class = 20
frequency (f₀) of class preceding the modal class = 12
frequency (f₂) of class succeeding the modal class = 11
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=40+\left(\frac{20-12}{2 \times 20-12-11}\right) \times 10$
$=40+\frac{8}{17} \times 10$
= 40 + 4.7
= 44.7

Question 9

Find the mode of the following distribution:

Sol :
Here, the maximum class frequency is 50, and the class corresponding to this frequency is 16 – 20.
So, the modal class is 16 – 20.
Now, modal class = 16 – 20, lower limit (l) of modal class = 16, class size(h) = 4
frequency (f₁) of the modal class = 50
frequency (f₀) of class preceding the modal class = 30
frequency (f₂) of class succeeding the modal class = 40
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=16+\left(\frac{50-30}{2 \times 50-30-40}\right) \times 4$
$=16+\frac{20}{30} \times 4$
= 16 + 2.67
= 18.67

Question 10

Find the mode of the following distribution:

Sol :

Here, the maximum number of persons i.e. 30 have income in the interval 300 – 400.
So, the modal class is 300 – 400.
lower limit (l) of modal class = 300,
class size(h) = 100
frequency (f₁) of the modal class = 30
frequency (f₀) of class preceding the modal class = 18
frequency (f₂) of class succeeding the modal class = 20
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=300+\left(\frac{30-18}{2 \times 30-20-18}\right) \times 100$
$=300+\frac{12}{60-38} \times 100$
= 300 + 54.54
= Rs. 354.54

Question 11

Find the mode of the following data:

Sol :

Here, the maximum no. of students is 9, and the class corresponding to the frequency is 20 – 30.
So, the modal class is 20 – 30.
lower limit (l) of modal class = 20,
class size(h) = 10
frequency (f₁) of the modal class = 9
frequency (f₀) of class preceding the modal class = 5
frequency (f₂) of class succeeding the modal class = 3
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=20+\left(\frac{9-5}{2 \times 9-5-3}\right) \times 10$
$=20+\frac{4}{10} \times 10$
= 20 + 4
= 24

Question 12

Find the mode of the following distribution:

Sol :

Here, the maximum no. of students is 20, and the class corresponding to this frequency is 20 – 30
So, the modal class is 20 – 30.
lower limit (l) of modal class = 20,
class size(h) = 10
frequency (f₁) of the modal class = 20
frequency (f₀) of class preceding the modal class = 6
frequency (f₂) of class succeeding the modal class = 10
Now, let us substitute these values in the formula

Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=20+\left(\frac{20-6}{2 \times 20-6-10}\right) \times 10$
$=20+\frac{14}{24} \times 10$
= 20 + 5.83
= 25.83

Question 13

Find the mode of the following distribution:

Sol :

Here, the maximum frequency is 25, and the class corresponding to this frequency is 40 - 50
So, the modal class is 40 – 50.
lower limit (l) of modal class = 40,
class size(h) = 10
frequency (f₁) of the modal class = 25
frequency (f₀) of class preceding the modal class = 12
frequency (f₂) of class succeeding the modal class = 10
Now, let us substitute these values in the formula
Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=40+\left(\frac{25-12}{2 \times 25-12-10}\right) \times 10$
$=40+\frac{13}{28} \times 10$
= 40 + 4.64
Mode = 44.64(approx.)

Question 14

If the mode of the following distribution is Rs. 24, find the missing frequency:

Sol :
Given: Mode = Rs. 24
Frequency (f_i) = 100
Class size = 10
Let frequency of the class 10 – 20 be f₀. Then, the frequency of the class 30 – 40 will be equal to 100 – 14 – f₀ – 27 – 15 = 44 – f₀.
It is given that mode = 24, therefore, modal class is 20 – 30.
Thus, lower limit of the modal class (l) = 20
frequency (f₀) of class preceding the modal class = f₀
frequency (f₂) of class succeeding the modal class = 44 – f₀

Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$\Rightarrow 24=20+\left(\frac{27-f_{0}}{2 \times 27-f_{0}-44+f_{0}}\right) \times 10$
$\Rightarrow 24-20=\left(\frac{27-\mathrm{f}_{0}}{10}\right) \times 10$
⇒ 4 = 27 - f₀
⇒ -23 = -f₀
⇒ f₀ = 23
∴ f₂ = 44 – 23 = 21
Hence, the missing frequencies are 23 and 21.

Question 15

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:

Sol :

Now, $\overline{\mathrm{x}}=\mathrm{a}+\mathrm{h}\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{u}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right)$
$\Rightarrow \overline{\mathrm{x}}=37.5+5\left(\frac{-58}{35}\right)$
⇒x̄ = 37.5 – 8.28
⇒x̄ = 29.22
∴ Mean = 29.22

Here, the maximum no. of students per teacher i.e. 10, and the class corresponding to this frequency is 30 – 35
So, the modal class is 30 – 35 .
lower limit (l) of modal class = 30,
class size(h) = 5
frequency (f₁) of the modal class = 10
frequency (f₀) of class preceding the modal class = 9
frequency (f₂) of class succeeding the modal class = 3
Now, let us substitute these values in the formula

Mode $=1+\left(\frac{\mathrm{f}_{1}-\mathrm{f}_{0}}{2 \mathrm{f}_{1}-\mathrm{f}_{0}-\mathrm{f}_{2}}\right) \times \mathrm{h}$
$=30+\left(\frac{10-9}{2 \times 10-9-3}\right) \times 5$
$=30+\frac{1}{8} \times 5$
= 30 + 0.625
Mode = 30.625(approx.)

S.no	Chapters	Links
1	Real numbers	Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4
2	Polynomials	Exercise 2.1 Exercise 2.2 Exercise 2.3
3	Pairs of Linear Equations in Two Variables	Exercise 3.1 Exercise 3.2 Exercise 3.3 Exercise 3.4 Exercise 3.5
4	Trigonometric Ratios and Identities	Exercise 4.1 Exercise 4.2 Exercise 4.3 Exercise 4.4
5	Triangles	Exercise 5.1 Exercise 5.2 Exercise 5.3 Exercise 5.4 Exercise 5.5
6	Statistics	Exercise 6.1 Exercise 6.2 Exercise 6.3 Exercise 6.4
7	Quadratic Equations	Exercise 7.1 Exercise 7.2 Exercise 7.3 Exercise 7.4 Exercise 7.5
8	Arithmetic Progressions (AP)	Exercise 8.1 Exercise 8.2 Exercise 8.3 Exercise 8.4
9	Some Applications of Trigonometry: Height and Distances	Exercise 9.1
10	Coordinates Geometry	Exercise 10.1 Exercise 10.2 Exercise 10.3 Exercise 10.4
11	Circles	Exercise 11.1 Exercise 11.2
12	Constructions	Exercise 12.1
13	Area related to Circles	Exercise 13.1
14	Surface Area and Volumes	Exercise 14.1 Exercise 14.2 Exercise 14.3 Exercise 14.4
15	Probability	Exercise 15.1