Question

Let X be a random variable with distribution.

x	-2	-1	3	4	6
P(X = x)	$\frac{1}{5}$	a	$\frac{1}{3}$	$\frac{1}{5}$	b

If the mean X is 2.3 and variance of X is ${\sigma }^2,\text{then}100{\sigma }^2$ is equal to

Answer 1

$\frac{1}{5}+\frac{1}{3}+\frac{1}{5}+a+b=1\Rightarrow a+b=\frac{4}{15}$

$\sum P_i{X}_i=2.3\Rightarrow \frac{1}{5}(-2)-a+1+\frac{4}{5}+6b=\frac{23}{10}$

$\begin{array}{c}-a+6b=\frac{9}{10}\\ a+b=\frac{4}{15}\end{array}\}\begin{array}{c}a=\frac{1}{10}\\ b=\frac{1}{6}\end{array}$

$\text{Variance}=\sum P_i{X}_i^2-(\overline{X}{)}^2$

$=(\frac{1}{5}.4+a+3+\frac{16}{5}+36b)-{\left(\frac{23}{10}\right)}^2$

$=\frac{781}{100}$

$100{\sigma }^2=781$