Chapter 16, Problem 37E

(a)

To determine

To calculate: the amount of cereals in big bowl that are expecting.

Answer to Problem 37E

1 ounce

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured into a small bowl

Suppose Y is the sum in a big bowl of cereal.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Formula used:

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

Calculation:

It could expect Y-X more amount of cereal in the large bowl to be

  $\begin{array}{c}E\left(Y-X\right)=E\left(Y\right)-E\left(X\right)\\ =2.5-1.5\\ =1\text{ ounce}\end{array}$

(b)

To determine

To find: the standard deviation of this variation can be identified.

Answer to Problem 37E

0.5

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured into a small bowl

Suppose Y is the sum in a big bowl of cereal.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Formula used:

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

Calculation:

Finding the variance first

  $\begin{array}{c}E\left(Y-X\right)=\mathrm{var}\left(Y\right)+\mathrm{var}\left(X\right)\\ ={\left(0.4\right)}^2-{\left(0.3\right)}^2\\ =0.16+0.09\\ =0.25\end{array}$

  $\begin{array}{c}SD\left(Y-X\right)=\sqrt{\mathrm{var}\left(Y-X\right)}\\ =\sqrt{0.25}\\ =0.5\end{array}$

Standard deviation = 0.5

(c)

To determine

To find: the probability is that the small bowl holds more cereal than the big bowl.

Answer to Problem 37E

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured into a small bowl

Suppose Y is the sum in a big bowl of cereal.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Calculation:

Normal (1, 0.5)

Small bowl is having more cereal than large one is X − Y > 0

  $\begin{array}{c}D=Y-X\\ Z=\frac{0-1}{0.5}\\ =-2\\ P\left(D<0\right)=P\left(Z<-2\right)\\ =0.02275\end{array}$

Required Probability is 0.02275

(d)

To determine

To find: the mean and standard deviation of the overall cereal quantity in the two bowls.

Answer to Problem 37E

Mean = 4

Standard Deviation = 0.5

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured into a small bowl

Suppose Y is the sum in a big bowl of cereal.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Formula used:

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

Calculation:

The total amount of cereal in the bowl is X+Y.

  $\begin{array}{c}E\left(X+Y\right)=E\left(X\right)+E\left(Y\right)\\ =1.5+2.5\\ =4\end{array}$

  $\begin{array}{c}\mathrm{var}\left(X+Y\right)=\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ ={\left(0.3\right)}^2+{\left(0.4\right)}^2\\ =0.25\\ SD\left(X+Y\right)=\sqrt{\mathrm{var}\left(X+Y\right)}\\ =\sqrt{0.25}\\ =0.5\end{array}$

(e)

To determine

To find: the possibility of over 4.5 ounces of cereal pouring together in the two bowls.

Answer to Problem 37E

0.1587

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured into a small bowl

Suppose Y is the sum in a big bowl of cereal.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Calculation:

Total follows normal mode $X+Y-N\left(4,0.5\right)$

Let D = X + Y

  $\begin{array}{c}P\left(D>4.5\right)=P\left(z>\frac{4.5-4}{0.5}\right)\\ =P\left(Z-1\right)\\ =0.1587\end{array}$

Required Probability is 0.1587

(f)

To determine

To find: the estimated cereal sum left in the box and the standard deviation.

Answer to Problem 37E

0.54

Explanation of Solution

Given:

Suppose X is the quantity of cereal that could be poured into a small bowl

Suppose Y is the sum in a big bowl of cereal.

	Mean	SD
X	1.5	0.3
Y	2.5	0.4

Formula used:

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

Calculation:

Suppose z is the amount of cereal put in the boxes.

  $\begin{array}{c}E\left(Z\right)=16.3\\ SD\left(Z\right)=0.2\end{array}$

Amount of cereal remaining is

  $\begin{array}{c}E\left(Z-\left(X+Y\right)\right)=E\left(Z\right)-E\left(X+Y\right)\\ =16.3-1.5-2.5\\ =12.3\\ \mathrm{var}\left(Z-\left(X+Y\right)\right)=\mathrm{var}\left(Z\right)+\mathrm{var}\left(X\right)+\mathrm{var}\left(Y\right)\\ ={\left(0.2\right)}^2+{\left(0.3\right)}^2+{\left(0.4\right)}^2\\ =0.04+0.09+0.16\\ =0.29\end{array}$

  $\begin{array}{c}SD\left(Z-\left(X+Y\right)\right)=\sqrt{\mathrm{var}\left(Z-\left(X+Y\right)\right)}\\ =\sqrt{0.29}\\ =0.54\end{array}$

The estimated cereal in the box is 12.3 ounces at a standard deviation of 0.54.