Chapter 12.2, Problem 64SR

a.

To determine

Find the percent of the CDs that would have greater than $120mm$ .

Answer to Problem 64SR

The percentage of the CDs will be is $50\%$ .

Explanation of Solution

Given: It is given in the question that the sizes of CDs made by the company are normally distributed with a standard deviation of 1 mm. The CDs are supposed to be $120mm$ in diameter, and they are made for drives that are $122mm$ wide.

Concept Used:

In this, use the concept of standard deviation and mean method and some knowledge of percentage.

Calculation:

Since, based on the set up of standard deviation and mean,

  $120mm$ is the mean , so $50\%$ would be greater.

Conclusion:

The percentage is $50\%$ .

b.

To determine

Find the number of CDs made in one hour which is between $119mm\&122mm$ .

Answer to Problem 64SR

The number of CDs is $815CDs$ .

Explanation of Solution

Given: It is given in the question that the sizes of CDs made by the company are normally distributed with a standard deviation of 1 mm. The CDs are supposed to be $120mm$ in diameter, and they are made for drives that are $122mm$ wide.

Concept Used:

In this, use the concept of standard deviation and mean method and some knowledge of percentage.

Calculation: Since, based on the set up of standard deviation and mean,

  $68\%$ accounts for CDs between $119mm\&121mm$

and $95\%$ accounts for CDs between $118mm\&122mm$ .

So, the percentage for $119mm\&122mm$ :

  $\frac{95\%}{2}+\frac{68\%}{2}=47.5\%+34\%=81.5\%\approx 815CDs$

Conclusion:

The number of CDs is $815CDs$ .

c.

To determine

Find the number of CDs formed too large to fit in the drives.

Answer to Problem 64SR

The number of CDs are $25CDs$ .

Explanation of Solution

Given:

It is given in the question that the sizes of CDs made by the company are normally distributed with a standard deviation of 1 mm. The CDs are supposed to be $120mm$ in diameter, and they are made for drives that are $122mm$ wide.

Concept Used:

In this, use the concept of standard deviation and mean method and some knowledge of percentage.

Calculation: Based on the set up of standard deviation and mean, Half of all CDs are above the mean, $95\%$ of the CDs are within two standard deviations, so that means $95\%$ of CDs are between $118mm\&122mm$ . Just only want the CDs above $122mm$ , so that leaves that last $5\%$ , bit only half of that is at the top end of the curve, half is at the bottom end of the curve.

  $50\%-\frac{95\%}{2}=\frac{100\%-95\%}{2}=\frac{5\%}{2}=2.5\%\approx 25CDs$

Conclusion:

The number of CDs are $25CDs$ .