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# 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 + (Σfᵢuᵢ/Σfᵢ) × h

Modal Class is the class with the highest frequency

Mode = l + [(f₁ - f₀) / (2f₁ - f₀ - f₂)] × h

Here,

Class size, h

Lower limit of modal class, l

Frequency of modal class, f₁

Frequency of class preceding modal class, f₀

Frequency of class succeeding the modal class, f₂

Median Class is the class having Cumulative frequency(cf) just greater than n/2

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] × 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, Σfᵢ = 100 and Σfᵢuᵢ = -106

Class size, h = 3

Mean, (x) = a + (Σfᵢuᵢ/Σfᵢ) × 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, f₁ = 40

Frequency of class preceding modal class, f₀ = 30

Frequency of class succeeding the modal class, f₂ = 16

Mode = l + [(f₁ - f₀)/(2f₁ - f₀ - f₂)] × 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.

**☛ Check: **NCERT Solutions for Class 10 Maths Chapter 14

**Video Solution:**

## 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

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

**Summary:**

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. The median and mean number of letters in the 100 surnames randomly picked up are 8.05 and 8.32 respectively while the modal size of surnames is 7.88.

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