NCERT Solutions class 10 maths chapter 14

Statistics

NCERT Solutions class 10 maths chapter 14 Exercise 14.2

Question 1

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 1

greatest frequency f₁ = 23 , modal class = 35 – 45
l = 35
class width h = 10
f_o = 23,
f₁ = 21 and f₂ = 14
TWe know
Mode =
Mode = 35+[ ]×10
Mode = 35+() = 35 + 1.8
Mode = 36.8 year
So the mode is = 36.8 year



Mean =
= = 35.37 years
the mean of the given data = 35.37

Question 2

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

Determine the modal lifetimes of the components

Answer 2

the modal class is 60–80. l = 60
f_o = 61, f₁ = 52, f₂ = 38 and h = 20 Mode =
we get
Mode =60+[ ]×20
Mode = 60+( )
Mode = 60+( ) = 60+ 5.625
modal lifetime of the components = 65.625 hours.

Question 3

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 3

Modal class = 1500-2000,
l = 1500
f_o = 40 f₁ = 24, f₂ = 33 and h = 500
We know
Mode =
we get
Mode =1500+[ ]×500
Mode = 1500+( )
Mode = 1500+( ) = 1500 + 347.83
modal monthly expenditure of the families = Rs 1847.83


Mean = x̄
= 2750+( )×500
= 2750-87.50
= 2662.50
mean monthly expenditure of the families = Rs 2662.50

Question 4

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 4

Modal class = 30 – 35
l = 30
Class width h = 5
f_o = 10, f₁ = 9 and f₂ = 3
We know
Mode =
Mode = 30+()×5
Mode = 30+() = 30+0.625
Mode = 30.625


Mean =
= = 29.2
mean = 29.2

Question 5

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

Answer 5

Modal class = 4000 – 5000
l = 4000
class width h = 1000
f_o = 18, f₁= 4 and f = 9
We know
Mode =
Mode = 4000+( )×1000
Mode = 4000+( ) = 4000+608.695
Mode = 4608.695

Question 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 6

Modal class = 40 – 50, l = 40
Class width h= 10, f_o = 20, f₁ = 12 and f₂ = 11
Mode =
Mode = 40+( )×10
Mode = 40 + ( ) = 40 + 4.7 = 44.7
Thus, the mode = 44.7