Question

100 surnames were randomly picked up from a local telephone directly and the frequency distribution of the number of letter 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
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.

Answer 1

SOLUTION :  

FREQUENCY DISTRIBUTION TABLE is in the attachment

For MODE :

Here the maximum frequency is 40, and the class corresponding to this frequency is 7 – 10. So the modal class is 7 - 10.

Therefore, l = 7, h = 3,  f1= 40,  f0= 30 , f2 = 16

Mode = l + [(f1- f0) / (2f1- f0 - f2)] ×h

= 7 + [(40 - 30)/(2 × 40 - 30 – 16) ] ×3

= 7 + [10×3)/(80 - 46)]

= 7 +[30/ 34]

= 7 + 0.88

= 7.88  

MODE = 7.88

Hence, the mode size of the surnames is  7.88 .

MEAN :  

From the table : Σfi = 100 , Σfixi = 832

Mean = Σfixi /Σfi

Mean =  832/100 = 8.32

Hence, the Mean number of letters in the surnames is 8.32 .

For MEDIAN :  

Here, n = 100

n/2 = 50

Since, the Cumulative frequency just greater than 50  is 76 and the corresponding class is 7 - 10 .  Therefore 7 - 10 is the median class.

Here, l = 7  , f = 40 , c.f = 36,  h = 3

MEDIAN = l + [(n/2 - cf )/f ] ×h

= 7 + [(50 - 36)/40] × 3

= 7 + [14/40)]× 3

= 7 +( 7 × 3) /20

= 7 + 21/20

= 7 + 1.05

= 8.05  

Median = 8.05  

Hence, the Median number of letters in the surnames is 8.1.

★★ Mode = l + (f1-f0/2f1-f0-f2) ×h

l = lower limit of the modal class

h = size of the class intervals

f1 = frequency of the modal class

f0 = frequency of the class preceding the modal class

f2 = frequency of the class succeed in the modal class.

★★ MEDIAN = l + [(n/2 - cf )/f ] ×h

Where,

l = lower limit of the median class

n = number of observations  

cf = cumulative frequency  of class interval preceding the  median class

f = frequency  of median class

h = class  size

HOPE THIS ANSWER WILL HELP YOU…

Answer 2

Answer:

i)

We know that the area of the rectangles are proportional to the frequencies in a histogram.

Here the widths of the rectangles are varying. So we need to make certain modification in the length of the rectangles, so that the areas are become proportional to the frequencies.

The minimum class size is 2.

Length of a rectangle (adjusted frequency)=(minimum class size / class size of this class) × frequency.

The modified table of given data is in the attachment.

And the histogram with varying width is in the attachment.

ii)

The class interval in which the maximum number of surnames lie is 6-8

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Hope this will help you....

Step-by-step explanation: