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An incomplete frequency distribution is given as follows:

Variable
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Total
Frequency
   12
   30
   ?
   65
    ?
   25
    18
228

Given that the median is 46 . The missing frequencies using the median formula are,

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x=30, y=40

x=33, y=45

x=20, y=100

x=24, y=36

answer is B.

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Detailed Solution

Given frequency table is,

Variable	10-20	20-30	30-40	40-50	50-60	60-70	70-80	Total
Frequency	12	30	?	65	?	25	18	228

Let, the missing frequency for the class 30 - 40 is x and the missing frequency for the class 50 - 60 is y.

Variable	Frequency (f)	Cumulative frequency(c.f.)
10-20	12	12
20-30	30	42
30-40	x	x+42
40-50	65	x+107
50-60	y	x+y+107
60-70	25	x+y+132
70-80	18	x+y+150
Total	228	x+y+150

The relation between x and y from the total frequency,

x + y + 150 = 228 ⇒ x + y = 228 − 150 ⇒ x + y = 78

......(i)
We know,

Median = l + N 2 − c . f f × h

l = Lower limit of the median class
N = Number of observation
c.f .= Cumulative frequency of class, preceding the median class
f = frequency of the median class
h = class size.
We know,

N = x + y + 150

So,

N 2 = x + y + 150 2

It is given the median=46

∴ Median class = 40 − 50 l = 40, c . f . = x + 42, f = 65, h = 50 − 40 = 10 Median = 40 + x + y + 150 2 − (x + 42) 65 × 10

So,

⇒ 40 + x + y + 150 2 − (x + 42) 65 × 10 = 46

......(ii)
Solve the equation (i) and (ii) , we get,

⇒ 40 + 78 + 150 2 − (x + 42) 65 × 10 = 46 ⇒ 2282 − (x + 42) 65 × 10 = 46 − 40 ⇒ 114 − (x + 42) 65 × 10 = 6 ⇒ (72 − x) × 10 65 = 6 ⇒ 720 − 10 x = 390 ⇒ 10 x = 720 − 390 ⇒ x = 33010 ⇒ x = 33

Putting the value x = 33 on equation (i), we get,

33 + y = 78 ⇒ y = 78 − 33 ⇒ y = 45

Therefore, the missing frequency for the class 30 - 40 is 33 and the missing frequency for the class 50 - 60 is 45.
Hence, option 2 is correct.

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