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# 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.

 Class Interval Frequency Cumulative Frequency 1-4 6 6 4-7 30 36 7-10 40 76 10-13 16 92 13-16 4 96 16-19 4 100

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,
 Class Interval $$f_i$$ $$x_i$$ $$f_i x_i$$ 1-4 6 2.5 15 4-7 30 5.5 165 7-10 40 8.5 340 10-13 16 11.5 184 13-16 4 14.5 51 16-19 4 17.5 70 $$\sum f_i \ = \ 100$$ $$\sum f_i x_i \ = \ 825$$

$$\therefore$$ 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