Chapter 7, Problem 34E

a.

To determine

Find the expected number of muffler installations at City M’s branch that would take more than 30 minutes.

Answer to Problem 34E

The expected number of muffler installation at City M’s branch is 10.

Explanation of Solution

It is given that 20% of the mufflers are installed in more than 30 minutes.

The total number of mufflers installed by City M’s branch is 50. The number of muffler installation at City M’s branch follows the binomial distribution with $n=50$ and $\pi =0.20$.

The mean can be obtained as follows:

$\begin{array}{c}\mu =n\pi \\ =50\left(0.20\right)\\ =10\end{array}$

Therefore, the expected number of muffler installation at City M’s branch is 10.

The standard deviation can be obtained as follows:

$\begin{array}{c}\sigma =\sqrt{n\pi \left(1-\pi \right)}\\ =\sqrt{50\left(0.20\right)\left(1-0.20\right)}\\ =\sqrt{50\left(0.20\right)\left(0.80\right)}\\ =\sqrt{8}\\ =2.82\end{array}$.

b.

To determine

Find the likelihood that fewer than eight installations took more than 30 minutes.

Answer to Problem 34E

The likelihood that fewer than eight installations took more than 30 minutes is 0.1894.

Explanation of Solution

The likelihood that fewer than eight installations took more than 30 minutes can be obtained as follows:

$\begin{array}{c}P\left(X<8\right)=P\left(X<8-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X<7.5\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{7.5-10}{2.82}\right)\\ =P\left(Z<\frac{-2.5}{2.82}\right)\\ =P\left(Z<-0.88\right)\end{array}$

Step-by-step procedure to obtain the probability using Excel:

• Click on the Formulas tab in the top menu.
• Select Insert function. Then from category box, select Statistical and below that NORM.S.DIST.
• Click Ok.
• In the dialog box, Enter Z value as –0.88.
• Enter Cumulative as TRUE.
• Click Ok, the answer appears in the spreadsheet.

Output obtained using Excel is represented as follows:

From the above output, the probability of Z less than –0.88 is 0.1894.

Therefore, the likelihood that fewer than eight installations took more than 30 minutes is 0.1894.

c.

To determine

Find the likelihood that eight or fewer installations took more than 30 minutes.

Answer to Problem 34E

The likelihood that eight or fewer installations took more than 30 minutes is 0.2981.

Explanation of Solution

The likelihood that eight or fewer installations took more than 30 minutes can be obtained as follows:

$\begin{array}{c}P\left(X\le 8\right)=P\left(X<8+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X<8.5\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{8.5-10}{2.82}\right)\\ =P\left(Z<\frac{-1.5}{2.82}\right)\\ =P\left(Z<-0.53\right)\end{array}$

Step-by-step procedure to obtain the probability using Excel:

• Click on the Formulas tab in the top menu.
• Select Insert function, then from category box, select Statistical and below that NORM.S.DIST.
• Click Ok.
• In the dialog box, Enter Z value as –0.53.
• Enter Cumulative as TRUE.
• Click Ok, the answer appears in the spreadsheet.

Output obtained using Excel is represented as follows:

From the above output, the probability of Z less than –0.53 is 0.2981.

Therefore, the likelihood that eight or fewer installations took more than 30 minutes is 0.2981.

d.

To determine

Find the likelihood that exactly 8 of the 50 installations took more than 30 minutes.

Answer to Problem 34E

The likelihood that exactly 8 of the 50 installations took more than 30 minutes is 0.1087.

Explanation of Solution

The likelihood that exactly 8 of the 50 installations took more than 30 minutes can be obtained as follows:

$\begin{array}{c}P\left(X=8\right)=P\left(8-0.5<X<8+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(7.5<X<8.5\right)\\ =P\left(\frac{7.5-10}{2.82}<\frac{X-\mu }{\sigma }<\frac{8.5-10}{2.82}\right)\\ =P\left(\frac{-2.5}{2.82}<Z<\frac{-1.5}{2.82}\right)\\ =P\left(-0.88<Z<-0.53\right)\\ =P\left(Z<-0.53\right)-P\left(Z<-0.88\right)\end{array}$

From the previous Subpart b, the probability of Z less than –0.88 is 0.1894.

From the previous Subpart c, the probability of Z less than –0.53 is 0.2980.

Now, consider

$\begin{array}{c}P\left(X=8\right)=P\left(Z<-0.53\right)-P\left(Z<-0.88\right)\\ =0.2981-0.1894\\ =0.1087\end{array}$

Therefore, the likelihood that exactly 8 installations took more than 30 minutes is 0.1087.