Question

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?

Answer 1

`E(x) = 1.52+.5y`
`sigma(x) =sqrt(3.79136 +.125y^2)`

Explanation:

the expected value of x in the discrete case is

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

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

`E(x) = 0*.16 + 1*.04+ 2*.24 +5*.2 + y*.5= 1.52+.5y`

`sigma(x) =sqrt( (0-0*.16)^2 .16 + (1-1*.04)^2 .04+ (2-2*.24)^2 .24 + (5-5*.2)^2 *.2+ (y - .5y)^2 .5)`

`sigma(x) =sqrt( (.96)^2 .04+ (1.52)^2 .24 + (5-5*.2)^2 *.2+(.5y)^2 .5)`

`sigma(x) =sqrt(3.79136 +.125y^2)`