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The mean of the following frequency distribution is 57.6 and the sum of the observation is 50, then the missing frequencies $f 1$ and $f 2$ are,

Class Interval
0-20
20-40
40-60
60-80
80-100
100-120
Frequency
7
8
$f 1$
10
$f 2$
5

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8,12

12,12

8,8

2,10

answer is A.

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Detailed Solution

Given table is,

Class Interval	0-20	20-40	40-60	60-80	80-100	100-120
Frequency	7	8	$f 1$	10	$f 2$	5

Consider the table,

Class Interval	Frequency	$x i$	$f i x i$
0-20	7	10	70
20-40	8	30	240
40-60	$f 1$	50	50 $f 1$
60-80	10	70	700
80-100	$f 2$	90	90 $f 2$
100-120	5	110	550
	Sum= $30 + f 1 + f 2$

According to the given data,

∑ f i = 7 + 8 + f 1 + 10 + f 2 + 5 ⇒ ∑ f i = 30 + f 1 + f 2

The sum of

f i x i

is:

∑ f l x i = 70 + 240 + 50 f 1 + 700 + 90 f 2 + 550 ⇒ ∑ f l x i = 1560 + 50 f 1 + 90 f 2

The sum of frequency is,

30 + f 1 + f 2

Therefore,

N = 30 + f 1 + f 2 ⇒ 50 = 30 + f 1 + f 2 ⇒ f 1 + f 2 = 50 − 30 ⇒ f 1 + f 2 = 20 … … (1)

Equate the formula of mean with the given mean. Using the formula of mean we have:

x ¯ = ∑ f 1 x i ∑ f i ⇒ x ¯ = 1560 + 50 f 1 + 90 f 2 50

But we already know that the mean is 57.6.
Thus we will put the value of this mean in the equation we acquired.

57.6 = 1560 + 50 f 1 + 90 f 2 50 ⇒ 57.6 × 50 = 1560 + 50 f 1 + 90 f 2 ⇒ 2880 = 1560 + 50 f 1 + 90 f 2 ⇒ 5 f 1 + 9 f 2 = 2880 − 1560 10 ⇒ 5 f 1 + 9 f 2 = 132.......... (2)

Equate equations (1) and (2) and solve them. We have the equation (1) as:

f 1 + f 2 = 20 ⇒ f 1 = 20 − f 2 … … … (1)

We have the equation (2) as:

5 f 1 + 9 f 2 = 132 … … .. (2)

Putting the value of (1) in (2) we get,

5 20 − f 2 + 9 f 2 = 132 ⇒ 100 − 5 f 2 + 9 f 2 = 132 ⇒ 4 f 2 = 32 ⇒ f 2 = 324

⇒ f 2 = 8

Putting the value of

f 2 = 8

in equation (1) we have,

f 1 = 20 − 8 ⇒ f 1 = 12

The values of

f 1

and

f 2

are 12 and 8 respectively.
Therefore, option 1 is correct.

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