Chapter 7, Problem 33E

a.

To determine

Find the probability that she makes errors on more than six returns.

Answer to Problem 33E

The probability that she makes errors on more than six returns is 0.3483.

Explanation of Solution

It is given that an error is made on 7% of the returns prepared in the last year.

The total number of returns is 80. Thus, the distribution of error is binomial with $n=80$ and $\pi =0.07$.

The mean can be obtained as follows:

$\begin{array}{c}\mu =n\pi \\ =80\left(0.07\right)\\ =5.6\end{array}$

The standard deviation can be obtained as follows:

$\begin{array}{c}\sigma =\sqrt{n\pi \left(1-\pi \right)}\\ =\sqrt{80\left(0.07\right)\left(1-0.07\right)}\\ =\sqrt{80\left(0.07\right)\left(0.93\right)}\\ =\sqrt{5.208}\\ =2.28\end{array}$

The probability that she makes errors on more than six returns can be obtained as follows:

$\begin{array}{c}P\left(X>6\right)=P\left(X>6+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>6.5\right)\\ =P\left(\frac{X-\mu }{\sigma }>\frac{6.5-5.6}{2.28}\right)\\ =P\left(Z>\frac{0.9}{2.28}\right)\\ =P\left(Z>0.39\right)\\ =1-P\left(Z<0.39\right)\end{array}$

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 0.39.
• 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 0.39 is 0.6517.

Consider,

$\begin{array}{c}P\left(X>6\right)=1-P\left(Z<0.39\right)\\ =1-0.6517\\ =0.3483\end{array}$

Therefore, the probability that she makes errors on more than six returns is 0.3483.

b.

To determine

Find the probability that she makes errors on at least six returns.

Answer to Problem 33E

The probability that she makes errors on at least six returns is 0.516.

Explanation of Solution

The probability that she makes errors on at least six returns can be obtained as follows:

$\begin{array}{c}P\left(X\ge 6\right)=P\left(X>6-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>5.5\right)\\ =P\left(\frac{X-\mu }{\sigma }>\frac{5.5-5.6}{2.28}\right)\\ =P\left(Z>\frac{-0.1}{2.28}\right)\\ =P\left(Z>-0.04\right)\\ =1-P\left(Z<-0.04\right)\end{array}$

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 -0.04.
• 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 –0.04 is 0.4840.

Consider,

$\begin{array}{c}P\left(X\ge 6\right)=1-P\left(Z<-0.04\right)\\ =1-0.4840\\ =0.516\end{array}$

Therefore, the probability that she makes errors on at least six returns is 0.516.

c.

To determine

Find the probability that she makes errors on exactly six returns.

Answer to Problem 33E

The probability that she makes errors on exactly six returns is 0.1677.

Explanation of Solution

The probability that she makes errors on exactly six returns can be obtained as follows:

$\begin{array}{c}P\left(X=6\right)=P\left(6-0.5<X<6+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(5.5<X<6.5\right)\\ =P\left(\frac{5.5-5.6}{2.28}<\frac{X-\mu }{\sigma }<\frac{6.5-5.6}{2.28}\right)\\ =P\left(\frac{-0.1}{2.28}<Z<\frac{0.9}{2.28}\right)\\ =P\left(-0.04<Z<0.39\right)\\ =P\left(Z<0.39\right)-P\left(Z<-0.04\right)\end{array}$

From the previous Subpart a, the probability of Z less than 0.39 is 0.6517.

From the previous Subpart b, the probability of Z less than –0.04 is 0.4840.

Now, consider

$\begin{array}{c}P\left(X=6\right)=P\left(Z<0.39\right)-P\left(Z<-0.04\right)\\ =0.6517-0.4840\\ =0.1677\end{array}$

Therefore, the probability that she makes errors on exactly six returns is 0.1677.