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If the median of the following frequency distribution is $32.5$ . What are the missing frequencies?

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3, 6

4, 9

2, 8

None of these

answer is A.

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Detailed Solution

It is given the median of the following frequency distribution is

32.5

.
We first create a cumulative frequency table as shown below

Class Interval	Frequency(f)	Cumulative frequency
0 - 10	$f_{1}$	$f_{1}$
10 - 20	5	5+ $f_{1}$
20 - 30	9	14+ $f_{1}$
30 - 40	12	26+ $f_{1}$
40 - 50	$f_{2}$	26+ $f_{1}$ + $f_{2}$
50 - 60	3	29+ $f_{1}$ + $f_{2}$
60 -70	2	31+ $f_{1}$ + $f_{2}$
	N = 40

We know that, median is given as

Median = l + (\frac{\frac{n}{2} - cf}{f}) \times h

From the table, we get
Class Interval

h = 10

N = Σ f_{i} = 31 + f_{1} + f_{2}

\Rightarrow N = Σ f_{i} = 40

[given]

\Rightarrow 40 = 31 + f_{1} + f_{2}

\Rightarrow f_{2} = 9 - f_{1}

……[1]
Given median

= 32.5

\Rightarrow

Median class

= 30 - 40

\Rightarrow

Frequency of median class

f = 12

Lower boundary of median class

l = 30

Previous cumulative frequency of median class

cf = 14 + f_{1}

Putting the values in the formula, we get

32.5

= 30 + \frac{(\frac{31 + f_{1} + f_{2}}{2} - (14 + f_{1}))}{12} \times 10

\Rightarrow 32.5

= 30 + \frac{(3 - f_{1} + f_{2})}{12} \times 5

\Rightarrow 32.5

= \frac{360 + 15 - 5 f_{1} + 5 f_{2}}{12}

\Rightarrow 390 - 375 = - 5 f_{1} + 5 f_{2}

\Rightarrow - 5 f_{1} + 5 f_{2} = 15

\Rightarrow - f_{1} + f_{2} = 3

....(2)
On substituting (1) in (2), we get

- f_{1} + (9 - f_{1}) = 3

\Rightarrow - 2 f_{1} = - 6

\Rightarrow f_{1} = 3

Now, we substitute

f_{1} = 3

in (2), we get

- 3 + f_{2} = 3

{\Rightarrow f}_{2} = 6

Thus, the value of

f_{1} is 3

and

f_{2} is 6

.
Hence, the correct option is 1.

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