Chapter 6, Problem 30E

a)

To determine

To draw the model which shows 68-95-99.7

Explanation of Solution

Given:

  $\mu$ = 800

  $\sigma$ = 50

Following is the model which shows 68-95-99.7:

  

b)

To determine

To explain whether it is safe to use rivets a shear strength of 750 pounds.

Answer to Problem 30E

Yes, it is safe to use rivets a shear strength of 750 pounds.

Explanation of Solution

Given:

  $\mu$ = 800

  $\sigma$ = 50

  

As per rule of 68-95-99.7,

  $\mu -\sigma =800-50=750$

Using Normal model, it is safe to used rivets as 750 pounds.

c)

To determine

To find the percent of rivets would expect to fall below 900 pounds.

Answer to Problem 30E

Approximately 97.5% of the shear strengths should be below 900 pounds.

Explanation of Solution

Given:

  $\mu$ = 800

  $\sigma$ = 50

  

As per model, 900 is 2 standard deviation above the mean. Therefore, 95% of the data is within 2 standard deviation from the mean. Using symmetry, 5/2 = 2.5% is more than 2 standard deviations below the mean and 2.5% more than 2 standard deviations above the mean.

Hence, 2.5+95=97.5% of the shear strengths should be below 900 pounds.

d)

To determine

To find the maximum shear strength for which feel comfortable for approving this company’s rivets

Answer to Problem 30E

The maximum shear strength should be 900 pounds.

Explanation of Solution

Given:

  $\mu$ = 800

  $\sigma$ = 50

  

In general, the maximum usual range said to be $\mu +2\sigma$

That is, $\mu +2\sigma =800+2\times 50=900$

Hence, the maximum shear strength should be 900 pounds.