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100 surnames were randomly picked up from a local telephone directly 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 surnames: | 6 | 30 | 40 | 16 | 4 | 4 |
Given:
100 surnames were randomly picked up from a local telephone directly and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as given.
To do:
We have to determine the median number of letters in the surnames and the mean number of letters in the surnames and find the modal size of the surnames.
Solution:
The frequency of the given data is as given below.
Let the assumed mean be $A=8.5$.
We know that,
Mean $=A+\frac{\sum{f_id_i}}{\sum{f_i}}$
Therefore,
Mean $=8.5+(\frac{-18}{100})$
$=8.5-0.18$
$=8.32$
The mean of the given data is 8.32.
We observe that the class interval of 7-10 has the maximum frequency(40).
Therefore, it is the modal class.
Here,
$l=7, h=3, f=40, f_1=30, f_2=16$
We know that,
Mode $=l+\frac{f-f_1}{2 f-f_1-f_2} \times h$
$=7+\frac{40-30}{2 \times 40-30-16} \times 3$
$=7+\frac{10}{80-46} \times 3$
$=7+\frac{30}{34}$
$=7+0.88$
$=7.88$
The mode of the given data is 7.88.
Here,
$N=100$
This implies, $\frac{N}{2}=\frac{100}{2}=50$
Median class $=7-10$
We know that,
Median $=l+\frac{\frac{N}{2}-F}{f} \times h$
$=7+\frac{50-36}{40} \times 3$
$=7+\frac{42}{40}$
$=7+1.05=80.5$
The median of the given data is 8.05.
The mean, mode and median of the above data are 8.32, 7.88 and 8.05 respectively.
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