6.100 sumames were randomly picked up from a local telephone directory and the frequency distribution of the number cet letters in the English alphabet in the surnames was obtained as follows. NO. Of letters 1 - 4 4-7 7-10 10-13 13-16 16-19 No. of surnames 6 30 40 4 Find the mean number of letters in the surnames? Also, find the modal size of the surnames.
Answers
Answer:
Let us prepare the following table to compute the median :
Number of letters Number of surnames (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=n
We have, n=100
⇒
2
n
=50
The cumulative frequency just greater than
2
n
is 76 and the corresponding class is 7–10.
Thus, 7–10 is the median class such that
2
n
=50,l=7,f=40,cf=36 and h=3
Substitute these values in the formula
Median, M=l+
⎝
⎛
f
2
n
−cf
⎠
⎞
×h
M=7+(
40
50−36
)×3
M=7+
40
14
×3=7+1.05=8.05
Now, calculation of mean:
Number of letters Mid-Point (x
i
) Frequency (f
i
) f
i
x
i
1−4 2.5 6 15
4−7 5.5 30 165
7−10 8.5 40 340
10−13 11.5 16 184
13−16 14.5 4 58
16−19 17.5 4 70
Total 100 832
Therefore, Mean,
x
ˉ
=
∑f
i
∑f
i
x
i
=
100
832
=8.32
Calculation ofMode:
The class 7–10 has the maximum frequency therefore, this is the modal class.
Here,
l=7,h=3,f
1
=40,f
0
=30 and f
2
=16
Now, let us substitute these values in the formula
Mode =l+(
2f
1
−f
0
−f
2
f
1
−f
0
)×h
=7+
80−30−16
40−30
×3
=7+
34
10
×3=7+0.88=7.88
Hence, median =8.05, mean =8.32 and mode =7.88