Question

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Answer 1

CUMULATIVE FREQUENCY TABLE are in the attachment.  

Let the missing frequencies be  x and y.

Given : n(Σfi) = 200 , Mean = 1.46

From the table, Σfi = 86 + x + y , Σfixi = 140 + x + 2y

Σfi = 86 + x + y

200 = 86 + x + y

200 - 86 = x + y

x + y = 114

x = 114 - y …………(1)

MEAN = Σfixi/ Σfi

1.46 = (140 + x + 2y) / 200

1.46 × 200 = (140 + x + 2y)

292 = (140 + x + 2y)

x + 2y = 292 - 140

x + 2y = 152 ……….(2)

114 - y + 2y = 152

[From eq 1]

114 + y = 152

y = 152 - 114

y = 38

Put the value of y in eq 1  

x = 114 - y

x = 114 - 38

x = 76  

Hence, the missing frequencies be  x = 76 and y = 38 .

FOR MEDIAN :  

Here, n = 200

n/2 = 100

Since, the Cumulative frequency just greater than 100 is 122 and the value corresponding to 122 is 1.  

Hence, the Median is 1.  

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Answer 2

Hope it helps you...

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the first answer is correct.....