Chapter 10, Problem AP3.7CPT

To determine

The mean and the standers deviation of the total number of the silver wrappers.

1. $32,18.4$
2. $32,6.06$
3. $32,4.29$
4. $80,18.4$
5. $80,4.29$

Answer to Problem AP3.7CPT

The correct option is (c) $32,4.29$

Explanation of Solution

Given:

Holiday	$1$	$2$
Silver	$30\%$	$50\%$
Red	$30\%$	-
Pink	$40\%$	-
Blue	-	$50\%$

Randomly selected Pieces: $40$

Concept Used:

Standard deviation:

  $\text{SD}\left(X+Y\right)=\sqrt{\mathrm{var}\left(X+Y\right)}$

For the Distribution:

  $\text{Mean}\left(X+Y\right)=\text{Mean}\left(X\right)+\text{Mean}\left(Y\right)$

Calculation:

Each of the distribution a binomial distribution since it’s a yes or no whether not a silver candy was drawn. So let $X$ represents holiday $1$ binomial and $Y$ represent holiday $2$ . $X+Y$ Represent the total.

  $\begin{array}{l}X\sim binomial\left(40,0.3\right)\\ Y\sim binomial\left(40,0.5\right)\end{array}$

For any distribution regardless if they are individual and independent then the mean of the sum id just the sum of the means.

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

Since both of the holidays are independent then the variance of the total just the addition of the different variances.

  $\text{Var}\left(X+Y\right)=\text{Var}\left(X\right)+\text{Var}\left(Y\right)$

By getting the variance of the binomial is calculated to be $n\times p\times \left(1-p\right)$ .

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

The standard deviation is

  $\begin{array}{l}\text{SD}\left(X+Y\right)=\sqrt{\mathrm{var}\left(X+Y\right)}\\ \text{SD}\left(X+Y\right)=\sqrt{18.4}\\ \text{SD}\left(X+Y\right)=4.29\end{array}$

Conclusion:

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