Question

Give the standard deviation for the probability distribution. (round to two decimal places)

Answer 1

Firstly
`\text{Var}[X]=E[X^2]-(E[X])^2 \\ E[X^2]=\sum x^2\Pr(X=x) \\ E[X] = \sum x\Pr(X=x)`.

So work out E[X],

`E[X]=(1*0.20)+(2*0.30) + (3*0.10) + (4* 0.20)+(5*0.05)`, note we can get rid of the 0 because 0 times anything = 0

Similarly, work out E[X²]

`E[X]=(1^2*0.20)+(2^2*0.30) + (3^2*0.10) + (4^2* 0.20)+(5^2*0.05)`, note we can get rid of the 0 because 0 times anything = 0.

Then Var[X] = 6.75 - 2.15² = 2.1275
By using
`\text{SD}[X] = \sqrt{\text{Var}[X]}` you find SD=[X]=1.46