# Class 10 NCERT Solutions- Chapter 14 Statistics – Exercise 14.3

• Last Updated : 03 Mar, 2021

### Question 1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers in a locality. Find the median, mean, and mode of the data and compare them.

Solution:

Total number of consumer n = 68

n/2 =34

So, the median class is 125-145 with cumulative frequency = 42

Here, l = 125, n = 68, Cf = 22, f = 20, h = 20

Now we find the median:

Median =

= 125 + 12 = 137

Hence, the median is 137

Now we find the mode:

Modal class = 125 – 145,

Frequencies are

f1 = 20, f0 = 13, f2 = 14 & h = 20

Mode

On substituting the values in the given formula, we get

Mode =

= 125 + 140/13

= 125 + 10.77

= 135.77

Hence, the mode is 135.77

Now we find the mean:

= 135 + 20(7/68)

= 137.05

Hence, the mean is 137.05

Now, on comparing the median, mean, and mode, we found that mean, median and mode are more/less equal in this distribution.

### Question 2. If the median of a distribution given below is 28.5 then, find the value of x & y.

Solution:

According to the question

The total number of observations are n = 60

Median of the given data = 28.5

n/2 = 30

Median class is 20 – 30 with a cumulative frequency = 25 + x

Lower limit of median class, l = 20,

Cf = 5 + x,

f = 20 & h = 10

Now we find the median:

Median =

On substituting the values in the given formula, we get

28.5 =

8.5 = (25 – x)/2

17 = 25 – x

Therefore, x = 8

From the cumulative frequency, we can identify the value of x + y as follows:

60 = 5 + 20 + 15 + 5 + x + y

On substituting the values of x, we will find the value of y

60 = 5 + 20 + 15 + 5 + 8 + y

y = 60 – 53

y = 7

So the value of a is 8 and y is 7

### Question 3. The Life insurance agent found the following data for the distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to the persons whose age is 18 years onwards but less than the 60 years.

Solution:

According to the given question the table is

Given data: n = 100 and n/2 = 50

Median class = 35 – 45

Then, l = 35, cf = 45, f = 33 & h = 5

Now we find the median:

Median =

On substituting the values in the given formula, we get

Median =

= 35 + 5(5/33)

= 35.75

Hence, the median age is 35.75 years.

### Find the median length of leaves.

Solution:

The data in the given table are not continuous to reduce 0.5 in the lower limit and add 0.5 in the upper limit.

We get a new table:

From the given table

n = 40 and n/2 = 20

Median class = 144.5 – 153.5

l = 144.5,

cf = 17, f = 12 & h = 9

Now we find the median:

Median =

On substituting the values in the given formula, we get

Median =

= 144.5 + 9/4

= 146.75 mm

Hence, the median length of the leaves is 146.75 mm.

### Find the median lifetime of a lamp.

Solution:

According to the question

n = 400 and n/2 = 200

Median class = 3000 – 3500

l = 3000, Cf = 130,

f = 86 & h = 500

Now we find the median:

Median =

On substituting the values in the given formula, we get

Median =

= 3000 + 35000/86 = 3000 + 406.97

= 3406.97

Hence, the median lifetime of the lamps is 3406.97 hours

### Determine the number of median letters in the surnames. Find the number of mean letters in the surnames and also, find the size of modal in the surnames.

Solution:

According to the question

n = 100 and n/2 = 50

Median class = 7 – 10

Therefore, l = 7, Cf = 36, f = 40 & h = 3

Now we find the median:

Median =

On substituting the values in the given formula, we get

Median =

Median = 7 + 42/40 = 8.05

Hence, the median is 8.05

Now we find the mode:

Modal class = 7 – 10,

Where, l = 7, f1 = 40, f0 = 30, f2 = 16 & h = 3

Mode =

On substituting the values in the given formula, we get

Mode =

= 7 + 30/34 = 7.88

Hence, the mode is 7.88

Now we find the mean:

Mean =

= 825/100 = 8.25

Hence, the mean is 8.25

### Question 7. The distributions of below give a weight of 30 students of a class. Find the median weight of a student.

Solution:

According to the question

n = 30 and n/2 = 15

Median class = 55 – 60

l = 55, Cf = 13, f = 6 & h = 5

Now we find the median:

Median =

On substituting the values in the given formula, we get

Median =

= 55 + 10/6 = 55 + 1.666

= 56.67

Hence, the median weight of the students is 56.67

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