Chapter 16, Problem 39E

(a)

To determine

To Explain: the amount of cereal left in the box expected.

Answer to Problem 39E

12.2

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured in a little pan

Suppose that Y is the quantity of cereal in a large bowl.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Formula used:

  $E\left(T-\left(X+Y\right)\right)=E\left(T\right)-E\left(X\right)-E\left(Y\right)$

Calculation:

Expected value is

  $\begin{array}{c}E\left(T-\left(X+Y\right)\right)=E\left(T\right)-E\left(X\right)-E\left(Y\right)\\ =16.2-1.5-2.5\\ =12.2\end{array}$

The mean cereal in the box is 12.2

(b)

To determine

To find: the standard deviation.

Answer to Problem 39E

0.51

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured in a little pan

Suppose that Y is the quantity of cereal in a large bowl.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Formula used:

  $\begin{array}{c}\mathrm{var}\left(T-X-Y\right)=\mathrm{var}\left(T\right)+\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ SD\left(\left(T-\left(X+Y\right)\right)\right)=\sqrt{\mathrm{var}\left(T-\left(X+Y\right)\right)}\end{array}$

Calculation:

We first find the variance

  $\begin{array}{c}\mathrm{var}\left(T-X-Y\right)=\mathrm{var}\left(T\right)+\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ ={\left(0.1\right)}^2+{\left(0.3\right)}^2+{\left(0.4\right)}^2\\ =0.01+0.16+0.09\\ =0.26\\ SD\left(\left(T-\left(X+Y\right)\right)\right)=\sqrt{\mathrm{var}\left(T-\left(X+Y\right)\right)}\\ =\sqrt{0.26}\\ =0.51\end{array}$

Standard deviation is 0.15 Ounce.

(c)

To determine

To find: this box is more than 13 ounces possibly.

Answer to Problem 39E

5.8%

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured in a little pan

Suppose that Y is the quantity of cereal in a large bowl.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Calculation:

A normal model with mean 12.2 and standard deviation 0.51 is the left-over number L.

Required probability that greater than 13 ounces are

  $\begin{array}{c}P\left(L>13\right)=P\left(Z>\frac{13-12.2}{0.51}\right)\\ =P\left(Z>\frac{13-12.2}{0.51}\right)\\ =P\left(Z>1.57\right)\\ =0.058\end{array}$

The following graph gives the area of the probability

Graph:

The below graph mentions the area of the probability

  

Therefore, there is 5.8% possibility that box still is having greater than 13 ounces of cereal.