Question
Let X be a discrete random variable whose probability distribution is defined as follows:
$
P(X=x)=\left\{\begin{array}{cl}
k(x+1) & \text { for } x=1,2,3,4 \\
2 k x & \text { for } x=5,6,7 \\
0 & \text { otherwise }
\end{array}\right.
$
where k is a constant
Find:
i. k
ii. E(X)
iii. Standard deviation of X.

Answer

$
P(X=x)=\left\{\begin{array}{c}
k(x+1) \text { for } x=1,2,3,4 \\
2 k x \text { for } x=5,6,7 \\
0 \text { otherwise }
\end{array}\right.
$
Thus, we have following table:
X1234567otherwise
P(X)2k3k4k5k5k12k14k0
XP(X)2k6k12k20k20k72k98k0
X2P(X)2k12k36k80k80k432k686k0
add $X ^2 P ( X )$ in fourth row and 1st column
i. Since, $\sum P_i=1$
$
\Rightarrow K(2+3+4+5+10+12+14)=1 \Rightarrow k=\frac{1}{50}
$
ii. $\because E(X)=\sum X P(X)$
$
\begin{array}{l}
\therefore E(X)=2 k+6 k+12 k+20 k+50 k+72 k+98 k+0=260 k \\
=260 \times \frac{1}{50}=\frac{26}{5}=5.2\left[\because k=\frac{1}{50}\right] \ldots \text { (i) }
\end{array}
$
iii. We know that,
$\operatorname{Var}( X )=\left[ E \left( X ^2\right)\right]-[ E ( X )]^2=\sum X^2 P(X)-\left[\sum X P(X)\right]^2$
$=[2 k +12 k +36 k +80 k +250 k +432 k +686 k +0]-[5.2]^2 \ldots$ [using Eq. (i)$]$
$=[1498 k ]-27.04=\left[1498 \times \frac{1}{50}\right]-27.04\left[\because k=\frac{1}{50}\right]$
$=29.96-27.4=2.92$
We know that, standard deviation of $X =\sqrt{\operatorname{Var}(X)}=\sqrt{2.92}=1.7088=1.7$ (approx)

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