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. In separate random samples of $40$candies on Holiday $1$and $40$candies on Holiday $2$, what are the mean and standard deviation of the total number of silver wrappers?

a. $32,18.4$

b.$32,6.06$

c.$32,4.29$

d.$80,18.4$

e.$80,4.29$

We have to determine the mean and the standers deviation of the total number of the silver wrappers.

We will use the following concept for standard deviation and for distribution :

Standard deviation:

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

For the Distribution:

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

$$\begin{gathered} X~binomial(40,0.3) \\ Y~binomial(40,0.5) \end{gathered}$$

The sum of the means is the mean of every distribution, regardless of whether it is individual or independent.

$$\begin{gathered} Mean(X+Y)=Mean\left(X\right)+Mean\left(Y\right) \\ Mean\left(X\right)=40\times 0.3=12 \\ Mean\left(Y\right)=40\times 0.5=20 \\ Mean(X+Y)=32 \end{gathered}$$

Because the two holidays are unrelated, the overall variation is simply the sum of the individual variances.

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

The variance of the binomial is calculated as follows: $n\times p\times (1-p).$

$$\begin{gathered} Var\left(X\right)=40\times 0.3\times (1-0.3)=8.4 \\ Var\left(Y\right)=40\times 0.5\times 0.5=10 \\ Var(X+Y)=18.4 \end{gathered}$$

The standard deviation is

$$\begin{gathered} SD(X+Y)=\sqrt{var(X+Y)} \\ SD(X+Y)=\sqrt{18.4} \\ SD(X+Y)=4.29 \end{gathered}$$

Hence, the correct option is (c) $32,4.29$