Chapter 12.5, Problem 12PPS

a

To determine

To calculate: the percent of the product lasts between 150 and 210 days

Answer to Problem 12PPS

The required percent is $68.26\text{\%}$

Explanation of Solution

Given information:

The shelf life of a particular snack chip is normally distributed with a mean of 180 days and a standard deviation of 30 days.

Calculation:

Let $X$ is a random variable shows the shelf life of a particular snack chip. Here $x$ has normal distribution with following parameters,

  $\mu =180,\sigma =30$

(a)

The z-score for $x=150$ is,

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{150-180}{30}\end{array}$

  $=-1$

  $\begin{array}{l}\text{The z-score for }x=210\text{ is,}\\ z=\frac{x-\mu }{\sigma }\\ =\frac{210-180}{30}\\ =1\end{array}$

The percent of the product lasts between 150 and 210 days is,

  $P(150<x<210)=P(-1<z<1)$

  $=P(z\le 1)-P(z\le -1)$

  $=0.8413-0.1587$

  $=68.26\text{\%}$

Hence, the required percent is $68.26\text{\%}$

b

To determine

To calculate: the percent of the product lasts between 180 and 210 days

Answer to Problem 12PPS

The required percent is $34.13\text{\%}$

Explanation of Solution

Given information:

The shelf life of a particular snack chip is normally distributed with a mean of 180 days and a standard deviation of 30 days.

Calculation:

(b)

The z-score for $x=180$ is,

  $\begin{array}{cc}z& =\frac{x-\mu }{\sigma }\\ & =\frac{180-180}{30}\\ & =0\end{array}$

The z-score for $x=210$ is,

  $\begin{array}{cc}z& =\frac{x-\mu }{\sigma }\\ & =\frac{210-180}{30}\\ & =1\end{array}$

The percent of the product lasts between 180 and 210 days is,

  $P(180<x<210)=P(0<z<1)$

  $=P(z\le 1)-P(z\le 0)$

  $=0.8413-0.5$

  $=34.13\text{\%}$

Hence, the required percent is $34.13\text{\%}$

c

To determine

To calculate: the percent of the product lasts less than 90 days

Answer to Problem 12PPS

The required percent is $=0.13\text{\%}$

Explanation of Solution

Given information:

The shelf life of a particular snack chip is normally distributed with a mean of 180 days and a standard deviation of 30 days.

Calculation:

(c)

The z-score for $x=90$ is,

  $\begin{array}{cc}z& =\frac{x-\mu }{\sigma }\\ & =\frac{90-180}{30}\\ & =-3\end{array}$

The percent of the product lasts less than 90 days is,

  $P(x<90)=P(z<-3)$

  $=0.0013$

  $=0.13\text{\%}$

Hence, the required percent is $=0.13\text{\%}$

d

To determine

To calculate: the percent of the product lasts more than 210 days

Answer to Problem 12PPS

The required percent is $15.87\text{\%}$

Explanation of Solution

Given information:

The shelf life of a particular snack chip is normally distributed with a mean of 180 days and a standard deviation of 30 days.

Calculation:

(d)

The z-score for $x=210$ is,

  $\begin{array}{cc}z& =\frac{x-\mu }{\sigma }\\ & =\frac{210-180}{30}\\ & =1\end{array}$

The percent of the product lasts more than 210 days is,

  $P(x>210)=P(z>1)$

  $=1-P(z\le 1)$

  $=1-0.8413$

  $=15.87\text{\%}$

Hence, the required percent is $15.87\text{\%}$