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 :

Number of letters	Number of surnames
1 - 4	6
4 - 7	30
7 - 10	40
10 - 13	16
13 - 16	4
16 - 19	4

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.

We will use direct method to find the mean.

By formula,

Mean = $\dfrac{Σf$ixi}{Σf_i} = \dfrac{832}{100} = 8.32

Cumulative frequency distribution table is as follows :

Here, n = 100, $\dfrac{n}{2} = \dfrac{100}{2}$ = 50.

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

∴ 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

By formula,

Median = $l + \Big(\dfrac{\dfrac{n}{2} - cf}{f}\Big) \times h$

Substituting values we get :

$\text{Median} = 7 + \dfrac{50 - 36}{40} \times 3 \\[1em] = 7 + \dfrac{14 \times 3}{40} \\[1em] = 7 + \dfrac{42}{40} \\[1em] = 7 + 1.05 \\[1em] = 8.05$

By formula,

Mode = l + $\Big(\dfrac{f$1 - f0}{2f1 - f0 - f_2}\Big) \times h

Here,

1. Class size is h.

2. The lower limit of modal class is l

3. The Frequency of modal class is f₁.

4. Frequency of class preceding modal class is f₀.

5. Frequency of class succeeding the modal class is f₂.

From table,

Class 7 - 10 has the highest frequency.

∴ It is the modal class.

∴ l = 7, f₁ = 40, f₀ = 30, f₂ = 16 and h = 3.

Substituting values we get :

$\text{Mode} = 7 + \Big(\dfrac{40 - 30}{2 \times 40 - 30 - 16}\Big) \times 3 \\[1em] = 7 + \dfrac{10}{80 - 46} \times 3 \\[1em] = 7 + \dfrac{30}{34} \\[1em] = 7 + 0.88 \\[1em] = 7.88$

Hence, mean = 8.32, median = 8.05 and mode = 7.88