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Question
Question 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:
Number of letters1−44−77−1010−1313−1616−19Number of surnames63040644 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.
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:
Number of letters1−44−77−1010−1313−1616−19Number of surnames63040644
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.
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Solution
We can find cumulative frequencies with their respective class intervals as below:
Number of lettersFrequency (fi)Cumulativefrequency1−4664−73030+6=367−104036+40=7610−131676+16=9213−16492+4=9616−19496+4=100Total(n)100
Now we may observe that cumulative frequency just greater than 76 is belonging to
class n2(i.e,1002=50) in interval 7−10.Median class=7−10
Lower limit l of median class = 7
Cumulative frequency cf of class preceding median class = 36
Frequency f of median class = 40
Class size h = 3
Median=1+(n2−cff)×h
=7+(50−3640)×3
=7+14×340
=8.05
Now we can find the class marks of given class intervals by using relation:
Class mark=upper class limit+lower class limit2
Taking 11.5 as assumed mean a we can find di, ui and fiui according to step deviation method as below.
Number of lettersNumber of surnamesxixi−aui=xi−a3fiui1−462.5−9−3−184−7305.5−6−2−607−10408.5−3−1−4010−131611.500013−16414.531416−19417.5628Total100 −106
∑fiui=−106
∑fi=100
Mean ¯x=a+(∑fiui∑fi)
=11.5+(−106100)×3
=11.5−3.18=8.32
We know that:
3 median = mode + 2 mean
3(8.05) = mode + 2(8.32)
24.15 - 16.64 = mode
7.51 = mode
So, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.51.
Number of lettersFrequency (fi)Cumulativefrequency1−4664−73030+6=367−104036+40=7610−131676+16=9213−16492+4=9616−19496+4=100Total(n)100
Now we may observe that cumulative frequency just greater than 76 is belonging to
class n2(i.e,1002=50) in interval 7−10.Median class=7−10
Lower limit l of median class = 7
Cumulative frequency cf of class preceding median class = 36
Frequency f of median class = 40
Class size h = 3
Median=1+(n2−cff)×h
=7+(50−3640)×3
=7+14×340
=8.05
Now we can find the class marks of given class intervals by using relation:
Class mark=upper class limit+lower class limit2
Taking 11.5 as assumed mean a we can find di, ui and fiui according to step deviation method as below.
Number of lettersNumber of surnamesxixi−aui=xi−a3fiui1−462.5−9−3−184−7305.5−6−2−607−10408.5−3−1−4010−131611.500013−16414.531416−19417.5628Total100 −106
∑fiui=−106
∑fi=100
Mean ¯x=a+(∑fiui∑fi)
=11.5+(−106100)×3
=11.5−3.18=8.32
We know that:
3 median = mode + 2 mean
3(8.05) = mode + 2(8.32)
24.15 - 16.64 = mode
7.51 = mode
So, median number and mean number of letters in surnames is 8.05 and 8.32 respectively while modal size of surnames is 7.51.
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