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

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

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.

We have n = 100
\(\Rightarrow \frac{n}{2} \ = \ 50 \)

The cumulative frequency just greater than \( \frac{n}{2} \) is 76 and the corresponding class is 7-10 .

Thus 7-10 is the median class such that

\( \frac{n}{2} \ = \ 50 , l = 7, cf = 36 , f = 40 and h = 3 \)

Median \( = \ l \ + \ ( \frac{ \frac{n}{2} \ - \ cf}{f} ) \ × \ h \)

\( = \ 7 \ + \ ( \frac{50 \ - \ 36}{40}) \ × \ 3 \)

\( = \ 7 \ + \ \frac{14}{40} \ × \ 3 \)

\(= \ 7 \ + \ 1.05 \ = \ 8.05 \)

Now, to calculate the mean,

\(\therefore \) Mean = \( \overline{x} \ = \ \frac{ \sum f_i x_i}{ \sum f_i} \)

\( = \ \frac{832}{100} \)

Therefore, Mean = \( 8.32 \)

Calculation of mode :

The class 7-10 has the maximum frequency therefore, this is the modal class.

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

Mode \( = \ l \ + \ ( \frac{f_1 \ - \ f_0}{2f_1 \ - \ f_0 \ - \ f_2}) \ × \ h \)

\(= \ 7 \ + \ \frac{40 \ - \ 30}{80 \ - \ 30 \ - \ 16} \ × \ 3 \)

\(= \ 7 \ + \ \frac{10}{34} \ × \ 3 \)

\( = \ 7 \ + \ 0.88 \ = \ 7.88 \)

\(\therefore \) Median = 8.05 , Mean = 8.32 and Mode = 7.88