91影视

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.

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

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: $n脳p脳(1鈭�p).$

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

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) $32,4.29$