# What is the expected value and standard deviation of X if P(X=0) = 0.16, P(X=1) = 0.4, P(X=2) = 0.24, P(X=5)=0.2?

Jul 17, 2016

$E \left(x\right) = 1.52 + .5 y$
$\sigma \left(x\right) = \sqrt{3.79136 + .125 {y}^{2}}$

#### Explanation:

the expected value of x in the discrete case is

$E \left(x\right) = \sum p \left(x\right) x$ but this is with $\sum p \left(x\right) = 1$ the distribution given here does not sum to 1 so I will assume some other value does exist and call it $p \left(x = y\right) = .5$

and standard deviation
sigma(x) = sqrt(sum (x-E(x))^2p(x)

$E \left(x\right) = 0 \cdot .16 + 1 \cdot .04 + 2 \cdot .24 + 5 \cdot .2 + y \cdot .5 = 1.52 + .5 y$

$\sigma \left(x\right) = \sqrt{{\left(0 - 0 \cdot .16\right)}^{2} .16 + {\left(1 - 1 \cdot .04\right)}^{2} .04 + {\left(2 - 2 \cdot .24\right)}^{2} .24 + {\left(5 - 5 \cdot .2\right)}^{2} \cdot .2 + {\left(y - .5 y\right)}^{2} .5}$

$\sigma \left(x\right) = \sqrt{{\left(.96\right)}^{2} .04 + {\left(1.52\right)}^{2} .24 + {\left(5 - 5 \cdot .2\right)}^{2} \cdot .2 + {\left(.5 y\right)}^{2} .5}$

$\sigma \left(x\right) = \sqrt{3.79136 + .125 {y}^{2}}$