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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:

6.   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:

    3.6

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.

Answers (1)

best_answer

Class

Number of

surnames f_i

Cumulative

Frequency

Class mark

x_i

 f_ix_i
1-4 6 6 2.5 15
4-7 30 36 5.5 165
7-10 40 76 8.5 340
10-13 16 92 11.5 184
13-16 4 96 14.5 51
16-19 4 100 17.5 70
 

 

\sum f_i = N

= 100

 

\sum f_ix_i

= 825

MEDIAN:
N= 100 \implies \frac{N}{2} = 50
\therefore Median class = 7-10; Lower limit, l = 7; 

Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
Median = l + \left (\frac{\frac{n}{2}-c.f}{f} \right ).W
\\ = 7+ \left (\frac{50-36}{40} \right ).3 \\ \\

= 8.05

Thus, the median of the data is 8.05

MODE:

The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 7-10
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency (f_1) of the modal class = 40; frequency (f_0) of class preceding the modal class = 30, frequency (f_2) of class succeeding the modal class = 16

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

\\ = 7 + \left(\frac{40-30}{2(40)-30-16} \right).3 \\ \\ = 125 + \frac{10}{34}.3

= 7.88

Thus, Mode of the data is 7.88

MEAN:

Mean, 
\overline x =\frac{\sum f_ix_i}{\sum f_i} 
= \frac{825}{100} = 8.25

Thus, the Mean of the data is 8.25

Posted by

HARSH KANKARIA

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