If the mean of the following probability distribution of a random variable X is then the variance of the distribution is

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Solution

Q.
If the mean of the following probability distribution of a random variable X :

X 0 2 4 6 8

P(X) $a$ $2 a$ $a + b$ $2 b$ $3 b$

is $\frac{46}{9},$ then the variance of the distribution is [2024]

1 $\frac{151}{27}$

2 $\frac{173}{27}$

3 $\frac{566}{81}$

4 $\frac{581}{81}$

Ans.
(3)

X 0 2 4 6 8

P(X) $a$ $2 a$ $a + b^{3}$ $2 b$ $3 b$

$Mean = \sum x_{i} P (x_{i})$

$\Rightarrow \frac{46}{9} = 0 + 4 a + 4 a + 4 b + 12 b + 24 b \Rightarrow \frac{46}{9} = 8 a + 40 b$

$\Rightarrow 36 a + 180 b = 23$ ...(i)

Also, $\sum_{i - 1}^{n} P_{i} = 1$

$\Rightarrow 4 a + 6 b = 1$ ...(ii)

On solving (i) and (ii), we get

$a = \frac{1}{12}, b = \frac{1}{9}$

Now, $σ^{2} = \sum x_{i}^{2} P (x_{i}) - (\sum x_{i} P {(x_{i}))}^{2}$

$= 0 + 4 \times 2 a + 16 (a + b) + 36 (2 b) + 64 (3 b) - {(\frac{46}{9})}^{2}$

$= 8 (a + 2 (a + b) + 9 b + 24 b) - {(\frac{46}{9})}^{2}$

$= 8 (3 a + 35 b) - {(\frac{46}{9})}^{2}$ $= 8 (\frac{3}{12} + \frac{35}{9}) - {(\frac{46}{9})}^{2}$

$= 8 (\frac{149}{36}) - {(\frac{46}{9})}^{2}$ $= \frac{566}{81}$

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