NCERT Solutions Class 10 Maths Chapter 14 Statistics Ex 14.2

NCERT Solutions Class 10 Maths 
Chapter – 14 (Statistics) 

The NCERT Solutions in English Language for Class 10 Mathematics Chapter – 14 Statistics Exercise 14.2 has been provided here to help the students in solving the questions from this exercise. 

Chapter : 14 Statistics

• NCERT Class 10 Maths Solution Ex – 14.1
• NCERT Class 10 Maths Solution Ex – 14.3
• NCERT Class 10 Maths Solution Ex – 14.4

Exercise – 14.2

1. The following table shows the ages of the patients admitted to 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.

Solution – To find out the modal class, let us the consider the class interval with high frequency.
Here, the greatest frequency = 23, so the modal class = 35 – 45,
Lower limit of modal class = l = 35,
class width (h) = 10,
f_m = 23,
f₁ = 21 and f₂ = 14
The formula to find the mode is
Mode = l + [(f_m – f₁)/(2f_m – f₁ – f₂)] × h
Mode = 35 + [(23 – 21)/(46 – 21 – 14)]×10
= 35 + (20/11)
= 35 + 1.8
= 36.8 years
So the mode of the given data = 36.8 years
Calculation of Mean : First find the midpoint using the formula,
x_i = (upper limit + lower limit)/2

The mean formula is
Mean = x̄ = ∑f_ix_i / ∑f_i
= 2830/80
= 35.375 years
Therefore, the mean of the given data = 35.375 years

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

Determine the modal lifetimes of the components.
Solution – From the given data the modal class is 60–80.
Lower limit of modal class = l = 60,
The frequencies are:
f_m = 61,
f₁ = 52,
f₂ = 38
h = 20
The formula to find the mode is
Mode = l+ [(f_m – f₁)/(2f_m – f₁ – f₂)] × h
Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20
Mode = 60 + [(9 × 20)/32]
Mode = 60 + (45/8) = 60 + 5.625
Therefore, modal lifetime of the components = 65.625 hours.

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:

Solution –  Given data:
Modal class = 1500 – 2000,
l = 1500,
Frequencies:
f_m = 40,
f₁ = 24,
f₂ = 33,
h = 500
Mode formula :
Mode = l + [(f_m – f₁)/ (2f_m – f₁ – f₂)] × h
Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500
Mode = 1500 + [(16 × 500)/23]
Mode = 1500 + (8000/23) = 1500 + 347.83
Therefore, modal monthly expenditure of the families = Rupees 1847.83
Calculation for mean : First find the midpoint using the formula,
x_i =(upper limit + lower limit)/2
Let us assume a mean, (a) be 2750.

The formula to calculate the mean,
Mean = x̄ = a +(∑f_iu_i /∑f_i) × h
= 2750 + (-35/200) × 500
= 2750 – 87.50
= 2662.50
So, the mean monthly expenditure of the families = Rs. 2662.50

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

Solution –  Given data:
Modal class = 30 – 35,
l = 30,
Class width (h) = 5,
f_m = 10,
f₁ = 9
f₂ = 3
Mode Formula:
Mode = l + [(f_m – f₁)/ (2f_m – f₁ – f₂)] × h
Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5
= 30 + (5/8)
= 30 + 0.625
= 30.625
Therefore, the mode of the given data = 30.625
Calculation of mean : Find the midpoint using the formula,
x_i =(upper limit + lower limit)/2

Mean = x̄ = ∑f_ix_i /∑f_i
= 1022.5/35
= 29.2 (approx)
Therefore, mean = 29.2

5. 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.

Solution –  Given data :
Modal class = 4000 – 5000,
l = 4000,
class width (h) = 1000,
f_m = 18,
f₁ = 4
f₂ = 9
Mode Formula:
Mode = l + [(f_m – f₁)/ (2f_m – f₁ – f₂)] × h
Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000
= 4000 + (14000/23)
= 4000 + 608.695
= 4608.695
= 4608.7 (approximately)
Thus, 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 summarized it in the table given below. Find the mode of the data:

Solution –  Given Data :
Modal class = 40 – 50,
l = 40,
Class width (h) = 10,
f_m = 20,
f₁ = 12
f₂ = 11
Mode = l + [(f_m – f₁)/(2f_m – f₁ – f₂)] × h
Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10
= 40 + (80/17)
= 40 + 4.7
= 44.7
Thus, the mode of the given data is 44.7 cars.

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