# 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames? Also, find the modal size of the surnames

**Solution:**

We will find the mean by step-deviation method

Mean, (x) = a + Σf_{i}d_{i-}/ Σf_{i}

Modal Class is the class with the highest frequency

Mode = l + (f_{1} - f_{0})/(2f_{1} - f_{0} - f_{2}) × h

Here, Class size, h

Lower limit of modal class, l

Frequency of modal class, f_{1}

Frequency of class preceding modal class, f_{0}

Frequency of class succeeding the modal class, f_{2}

Median Class is the class having Cumulative frequency(cf) just greater than n

Median = l + (n/2 - cf)/f × h

Class size, h

Number of observations, n

Lower limit of median class, l

Frequency of median class, f

Cumulative frequency of class preceding median class, cf

Let us find the median.

From the table, it can be observed that n = 100 ⇒ n/2 = 50

Cumulative frequency (cf) just greater than 50 is 76, belonging to class 7 - 10.

Therefore, median class = 7 - 10

Class size, h = 3

Lower limit of median class, l = 7

Frequency of median class, f = 40

Cumulative frequency of class preceding median class, cf = 36

Median = l + (n/2 - cf)/f × h

= 7 + (50 - 36)/40 x 3

= 7 + 14/40 × 3

= 7 + 21/20

= 7 +1.05

= 8.05

Now, we will find the mean.

Class mark, x_{i} = (Upper class limit + Lower class limit)/2

Taking assumed mean, a = 11.5

From the table, we obtain, Σ f_{i} = 100 and Σ f_{i}u_{i} = -106

Class size, h = 3

Mean, (x) = a + (Σf_{i}u_{i}/ Σf_{i})× h

= 11.5 + (- 106/100) × 3

= 11.5 - 318/100

= 11.5 - 3.18

= 8.32

Let's find the mode.

From the table, it can be observed that the maximum class frequency is 40, belonging to class interval 7 − 10.

Class size, h = 3

Modal class = 7 − 10

Lower limit of modal class, l = 7

Frequency of modal class, f_{1} = 40

Frequency of class preceding modal class, f_{0} = 30

Frequency of class succeeding the modal class, f_{2} = 16

Mode = l + (f_{1} - f_{0})/(2f_{1} - f_{0} - f_{2}) × h

= 7 + [(40 - 30)/(2 × 40 - 30 - 16)] × 3

= 7 + (10/34) × 3

= 7 + 15/17

= 7 + 0.88

= 7.88

Therefore, the median and mean number of letters in surnames is 8.05 and 8.32 respectively and modal size of surnames is 7.88.

**Video Solution:**

### NCERT Solutions for Class 10 Maths - Chapter 14 Exercise 14.3 Question 6:

The median and mean number of letters in the 100 surnames randomly picked up from a local telephone directory are 8.05 and 8.32 respectively while the modal size of surnames is 7.88