Chapter 12.5, Problem 13PPS

a.

To determine

Find the percent of time to get more than $8ounces$ of coffee.

Answer to Problem 13PPS

  $50\%$ .

Explanation of Solution

Given information:

A vending machine dispenses about $8ounces$ of coffee. It normally distributed with a standard deviation of $0.3ounces$ .

Calculation:

Let $x$ is a random variable shows the amount of vending machine dispenses of coffee.

  $\mu =8,\sigma =0.3$

  $\begin{array}{l}z-scoreforx=8\\ z=\frac{x-\mu }{\sigma }\\ =\frac{8-8}{0.3}=0\\ thepercentottimetogetmorethan8ouncescoffee\\ P\left(x>8\right)=P\left(z>0\right)\\ =0.50\\ =50\%\end{array}$

Hence, the percent of time to get more than $8ounces$ of coffee is $50\%$ .

b.

To determine

Find the percent of time to get less than $8ounces$ of coffee.

Answer to Problem 13PPS

  $50\%$ .

Explanation of Solution

Given information:

A vending machine dispenses about $8ounces$ of coffee. It normally distributed with a standard deviation of $0.3ounces$ .

Calculation:

Let $x$ is a random variable shows the amount of vending machine dispenses of coffee.

  $\mu =8,\sigma =0.3$

  $\begin{array}{l}z-scoreforx=8\\ z=\frac{x-\mu }{\sigma }\\ =\frac{8-8}{0.3}=0\\ thepercentottimetogetlessthan8ouncescoffee\\ P\left(x<8\right)=P\left(z<0\right)\\ =0.50\\ =50\%\end{array}$

Hence, the percent of time to get less than $8ounces$ of coffee is $50\%$ .

c.

To determine

Find the percent of time to get $07.4and8.6$ ounce of coffee.

Answer to Problem 13PPS

  $95\%$ .

Explanation of Solution

Given information:

A vending machine dispenses about $8ounces$ of coffee. It normally distributed with a standard deviation of $0.3ounces$ .

Calculation:

Let $x$ is a random variable shows the amount of vending machine dispenses of coffee.

  $\mu =8,\sigma =0.3$

  $\begin{array}{l}z-scoreforx=7.4\\ z=\frac{x-\mu }{\sigma }\\ =\frac{7.4-8}{0.3}=-2\\ z-scoreforx=8.6\\ z=\frac{x-\mu }{\sigma }\\ =\frac{8.6-8}{0.3}=2\\ thepercentottimetoget7.4and8.6ouncescoffee\\ P\left(7.5<x<8.6\right)=P\left(-2<z<2\right)\\ =P\left(z\le 2\right)-P\left(z\le -2\right)\\ =0.9772-0.0228\\ =95.44\%\end{array}$

Hence, the percent of time to get $07.4and8.6$ ounces of coffee is $95\%$ .