Chapter 12.2, Problem 63SR

a.

To determine

To find: What percent of the CD’s would you expect to be greater than $120$ millimeters.

Answer to Problem 63SR

  $50\%$ of the CD’s we expect to be greater than $120$ millimeters.

Explanation of Solution

Given information:

  $\begin{array}{l}\mu =120\\ \sigma =1\end{array}$

CD’s are made for drives that are $122$ millimeters wide.

The percent of CDs would you expect to be greater than $120$ millimeters is

  $P\left(x>120\right)$

On solving

We get

  $\begin{array}{l}P\left(x>120\right)=34\%+13.5\%+2\%+0.5\%\\ \Rightarrow P\left(x>120\right)=50\%\end{array}$

Therefore,

  $50\%$ of the CD’s we expect to be greater than $120$ millimeters.

b.

To determine

To find: How many of the CDs made in one hour would you expect to be between $119$ and $122$ .

Answer to Problem 63SR

  $815$ CDs made in one hour would you expect to be between $119$ and $122$

Explanation of Solution

Given information:

  $\begin{array}{l}\mu =120\\ \sigma =1\end{array}$

Company manufactures $1000$ CD’s per hour

The probability of CDs that was made in one hour would be between $119$ and $122$ millimeters $\mu -\sigma$ and $\mu +2\sigma$ , that is between $120-1$ or $119$ and $120+2\left(1\right)$ or $122$ .

  $\begin{array}{l}P\left(119<x<122\right)=34\%+34\%+13.5\%\\ \Rightarrow P\left(119<x<122\right)=81.5\%\end{array}$

It means $81.5\%$ of $1000$ CDs $=815$ CD’s

Therefore,

  $815$ CDs made in one hour would you expect to be between $119$ and $122$ .

c.

To determine

To find: About how many CDs per hour will be too large to fit in the drives.

Answer to Problem 63SR

  $25$ CDs per hour will be too large to fit in the drives.

Explanation of Solution

Given information:

  $\begin{array}{l}\mu =120\\ \sigma =1\end{array}$

Company manufactures $1000$ CD’s per hour

The probability CDs that will be too large to fit in the drives is greater than $\mu +2\sigma$ , that is $120+2\left(1\right)$ or $122$ .

  $\begin{array}{l}P\left(x>122\right)=2\%+0.5\%\\ \Rightarrow P\left(x>122\right)=2.5\%\end{array}$

  $2.5\%$ of $1000$ CDs $=25$ CDs

Therefore,

  $25$ CDs per hour will be too large to fit in the drives.