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The height of $30$ boys of a class are given in the following table:

If by joining a boy of height $140$ cm, the median of the heights is changed from $M_{1}$ to $M_{2}$ , then $M_{1} - M_{2}$ , in cm is

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0.1

- 0.1

0

0.2

answer is C.

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Detailed Solution

Given data,
Question Image

Then,

Here

N = 30

\frac{N}{2} + 1

\Rightarrow 15 + 1

\Rightarrow 16

16

is under cumulative frequency

20

.
So, the median class will be

140 - 149

.
We get,

L_{1} = 139.5

L_{2} = 149.5

f = 10

n = 30

c = 10

Then the median will be,
Median

(M_{1}) = L_{1} + \frac{L_{2} - L_{1}}{f} (\frac{n}{2} - c)

\Rightarrow (M_{1}) = 139.5 + \frac{10}{10} (15 - 10)

\Rightarrow (M_{1}) = 139.5 + \frac{10}{10} \times 5

\Rightarrow (M_{1}) = 139.5 + 5

\Rightarrow (M_{1}) = 144.5

If we join a boy of height

140

cms, then

n = 31

f = 11

Median

M_{2} = 139.5 + \frac{149.5 - 139.5}{11} (15.5 - 10)

\Rightarrow (M_{2}) = 139.5 + \frac{10}{11} \times 5.5

\Rightarrow (M_{2}) = 139.5 + 10 \times 0.5

\Rightarrow (M_{2}) = 139.5 + 5

\Rightarrow (M_{2}) = 144.5

cms
Then,

M_{1} - M_{2} = 144.5 - 144.5

\Rightarrow M_{1} - M_{2} = 0

Thus, the value of

M_{1} - M_{2} = 0

.
Hence, the correct option is 3.

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