Chapter PIII, Problem 28RE

(a)

To determine

To find: When measured under a 900-pound load, the percentage of reverts chosen at random would break.

Answer to Problem 28RE

10.56%

Explanation of Solution

Given:

  $\begin{array}{l}\mu =950\\ \sigma =40\end{array}$

Formula used:

  $Z=\frac{X-\mu }{\sigma }$

Calculation:

suppose X follow a normal model with an average of 950 pounds and a standard deviation 40 pound deviation. suppose Z follow a typical model with an average of 0 pounds and a standard deviation 1-pound deviation.

Then $Z=\frac{X-950}{40}$

  $\begin{array}{c}P\left(X<900\right)=P\left(\frac{X-950}{40}<\frac{900-950}{40}\right)\\ =P\left(Z<\frac{900-950}{40}\right)\\ =P\left(Z<-1.25\right)\end{array}$

From z-table get that

  $P\left(Z<-1.25\right)=0.1056$

Therefore

  $P\left(X<900\right)=0.1056$

10.56% of rivets choose at arbitrary would break if tested under a 900-pound load.

(b)

To determine

Tofind: the number of reverts may required to test in order to estimate 3 that fail at money 900 pounds

Answer to Problem 28RE

29

Explanation of Solution

Proportion of the rivets that at the 900 pounds or below is $0.1056=\frac{3}{n}$ .

Therefore, sample size is estimated by

  $n=\frac{3}{0.1056}=28.4\text{ round up to 29}$