Question

A certain candy has different wrappers for various holidays. During Holiday $1$the candy wrappers are $30\%$silver, $30\%$red, and $40\%$pink. During Holiday $2$the wrappers are $50\%$silver and $50\%$blue. Forty pieces of candy are randomly selected from the Holiday $1$distribution, and $40$ pieces are randomly selected from the Holiday $2$ distribution. What are the expected value and standard deviation of the total number of silver wrappers?

Short answer

Result is:

$32,4.29$

Step-by-step solution

Given information

We have been give that

$X~binomial(40,0.3)Y~binomial(40,0.5)$

Each distributio follows a binomial distribution.

Simplify

For any distribution, if they are independent, the mean of the sum is just the sum of means.

$Mean(X+Y)=Mean\left(X\right)+Mean\left(Y\right)$

Mean of a binomial is calculated to$n*p$

$Mean\left(X\right)=40*0.3=12,Mean\left(Y\right)=40*0.5=20$

Since both holidays are independent

$Var(X+Y)=Var\left(X\right)+Var\left(Y\right)$

Variance for a binomial distributed is maximised

$Var\left(X\right)=40*0.3*(1-0.3)=8.4,Var\left(Y\right)=40*0.5*$

$0.5=10Var(X+Y)=18.4$

We will calculate standard deviation by

$SD(X+Y)=\sqrt{Var(X+Y)}=4.29$