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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 alphabet 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 |

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also

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 $=l+\frac{\frac{N}{2}-F}{f} \times h$

$=7+\frac{50-36}{40} \times 3$

$=7+\frac{42}{40}$

$=7+1.05=8.05$

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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