If the mean of the following probability distribution of a radam variable [X] : [X] 0 2 4 …

Question

If the mean of the following probability distribution of a radam variable $$\mathrm{X}$$ :

$$\mathrm{X}$$
    0
    2
    4
    6
    8

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

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

Accepted Answer

$X$
    0
    2
    4
    6
    8

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

$$\begin{aligned}
& \text { Mean }=\sum x_i P\left(x_i\right) \
& \frac{46}{9}=4 a+4 a+4 b+12 b+24 b \
& \frac{46}{9}=8 a+40 b \
& \frac{23}{9}=4 a+20 b \
& 36 a+180 b=23 \quad \text{.... (1)}
\end{aligned}$$

Sum of probability is 1

$$\Rightarrow 4 a+6 b=1 \quad \text{... (2)}$$

$$\begin{aligned}
& \text { Solving (1) and (2) } \
& a=\frac{1}{12}, b=\frac{1}{9} \
& \sigma^2=\sum x_i^2 P\left(x_i\right)-\left(\sum x_i P\left(x_i\right)\right)^2 \
& =4 \times 2 a+16(a+b)+36(2 b)+64(3 b)-\left(\frac{46}{9}\right)^2 \
& =8(a+2(a+b)+9 b+24 b)-\left(\frac{46}{9}\right)^2 \
& =8(3 a+35 b)-\left(\frac{46}{9}\right)^2 \
& =8\left(\frac{3}{12}+\frac{35}{9}\right)-\left(\frac{46}{9}\right)^2 \
& =8\left(\frac{149}{36}\right)-\left(\frac{46}{9}\right)^2=\frac{566}{81} \
\end{aligned}$$

Solutions · 1