Exercise 14.2 Page: 275

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.

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,

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

Substitute the values in the formula, we get

Mode = 35+[(23-21)/(46-21-14)]×10

Mode = 35+(20/11) = 35+1.8

Mode = 36.8 year

So the mode of the given data = 36.8 year

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

= 35.37 years

Therefore, the mean of the given data = 35.37 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.

l = 60,

The frequencies are:

f_m = 61, f₁ = 52, f₂ = 38 and h = 20

The formula to find the mode is

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the formula, we get

Mode =60+[(61-52)/(122-52-38)]×20

Mode = 60+((9 x 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 and

h = 500

Mode formula:

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the formula, we get

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

Substitute the values in the given formula

= 2750+(-35/200)×500

= 2750-87.50

= 2662.50

So, the mean monthly expenditure of the families = Rupees 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 and f₂ = 3

Mode Formula:

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values in the given formula

Mode = 30+((10-9)/(20-9-3))×5

Mode = 30+(5/8) = 30+0.625

Mode = 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

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 and f₂ = 9

Mode Formula:

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values

Mode = 4000+((18-4)/(36-4-9))×1000

Mode = 4000+(14000/23) = 4000+608.695

Mode = 4608.695

Mode = 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 and f₂ = 11

Mode = l+ [(f_m-f₁)/(2f_m-f₁-f₂)]×h

Substitute the values

Mode = 40+((20-12)/(40-12-11))×10

Mode = 40 + (80/17) = 40 + 4.7 = 44.7

Thus, the mode of the given data is 44.7 cars