# Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.2

• Last Updated : 03 Mar, 2021

### Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.

Solution:

The greatest frequency in the given table is 23, so the modal class = 35 – 45,

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l = 35,

Class width = 10, and the frequencies are

fm = 23, f1 = 21 and f2 = 14

Now, we find the mode using the given formula

Mode On substituting the values in the formula, we get

Mode = = 35 + (20/11) = 35 + 1.8

= 36.8

Hence, the mode of the given data is 36.8 year

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Mean = = ∑fixi /∑fi

= 2830/80

= 35.37 years

### Determine the modal lifetimes of the components.

Solution:

According to the given question

The modal class is 60 – 80

l = 60, and the frequencies are

fm = 61, f1 = 52, f2 = 38 and h = 20

Now, we find the mode using the given formula

Mode On substituting the values in the formula, we get

Mode =  = 60 + 45/8 = 60 + 5.625

Hence, the modal lifetime of the components is 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:

Solution:

According to the question

Modal class = 1500-2000,

l = 1500,and the frequencies are

fm = 40 f1 = 24, f2 = 33 and

h = 500

Now, we find the mode using the given formula

Mode On substituting the values in the formula, we get

Mode =  = 1500 + 8000/23 = 1500 + 347.83

So, the modal monthly expenditure of the families is 1847.83 Rupees

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Let us considered a mean, A be 2750

Mean = On substituting the values in the given formula = 2750 – 87.50

= 2662.50

Hence, the mean monthly expenditure of the families is  2662.50 Rupees

### 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

Solution:

According to the question

Modal class = 30 – 35,

l = 30,

Class width (h) = 5, and the frequencies are

fm = 10, f1 = 9 and f2 = 3

Now, we find the mode using the given formula

Mode On substituting the values in the formula, we get

Mode = = 30 + 5/8 = 30 + 0.625

= 30.625

Hence, the mode of the given data is 30.625

Now, we find the mean. So for that first we need to find the midpoint.

xi = (upper limit + lower limit)/2

Mean = = 1022.5/35

= 29.2

Hence, the mean is 29.2

### Find the mode of the data.

Solution:

According to the question

Modal class = 4000 – 5000,

l = 4000,

class width (h) = 1000, and the frequencies are

fm = 18, f1 = 4 and f2 = 9

Now, we find the mode using the given formula

Mode On substituting the values in the formula, we get

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

= 4608.695

Hence, the mode of the given data is 4608.7 runs

### Question 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:

According to the question

Modal class = 40 – 50, l = 40,

Class width (h) = 10, and the frequencies are

fm = 20, f1 = 12 and f2 = 11

Now, we find the mode using the given formula

Mode On substituting the values in the formula, we get

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

Hence, the mode of the given data is 44.7 cars

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