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. 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.$80,4.29$

Short answer

The mean and the standers deviation of the total number of the silver wrappers is$32,4.29$

Step-by-step solution

Given Information

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

Simplification

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

Standard deviation:

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

For the Distribution:

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

Each distribution is a binomial distribution since the outcome of pulling a silver candy is either yes or no.

Assume that $X$is the binomial of holiday $1$and $Y$is the binomial of holiday $2$.

$X+Y$

Representthewholeimage.

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

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) $32,4.29$