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Probability Part 2

Section: Mathematics

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Question : 17 of 100

Marks: +1, -0

If the mean of the following probability distribution of a random variable

X

X	0	2	4	6	8
P(X)	a	2a	a+b	2b	3b

\frac{46}{9}

, then the variance of the distribution is

[4 Apr 2024 Shift 2]

$\frac{581}{81}$
$\frac{566}{81}$
$\frac{173}{27}$
$\frac{151}{27}$

Solution: 👈: Video Solution

\sum P_i = 1

a + 2a + a + b + 2b + 3b = 1

4a + 6b = 1

. . . (I)

E(x) = \text{mean} = \frac{46}{9}

\sum P_i X_i = \frac{46}{9} \Rightarrow 4a + 4a + 4b + 12b + 24b = \frac{46}{9}

8a + 40b = \frac{46}{9}

4a + 20b = \frac{23}{9} \qquad \ldots \ldots \text{ (II)}

Subtract (I) from (II) we get

b = \frac{1}{9} \quad \text{and} \quad a = \frac{1}{12}

\text{Variance} = E(x_i^2) - E(x_i)^2

E(x_i^2) = 0^2 \times 9^2 + 2^2 \times 2a + 4^2(a + b) + 6^2(2b) + 8^2(3b)

= 24a + 280b

Put

a = \frac{1}{12} \qquad b = \frac{1}{9}

E(x_i^2) = 2 + \frac{280}{9} = \frac{298}{9}

\therefore \sigma^2 = E(x_i^2) - E(x_i)^2

= \frac{298}{9} - \left(\frac{46}{9}\right)^2

\sigma^2 = \frac{298}{9} - \frac{2116}{81}

= \frac{566}{81}

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