Chapter 7, Problem 36E

a.

To determine

Find the expected number of guests to accept the invitation.

Answer to Problem 36E

The expected number of guests to accept the invitation is 160.

Explanation of Solution

It is given that 80% of the guests respond to invitations that they will attend the wedding and the total number of invitations planned is 200. The number of guests responds to invitation that they will attend the wedding follows a binomial distribution.

That is, $n=200$ and $\pi =0.80$.

The mean can be obtained as follows:

$\begin{array}{c}\mu =n\pi \\ =200\left(0.80\right)\\ =160\end{array}$

Thus, the expected number of guests to accept the invitation is 160.

b.

To determine

Find the standard deviation.

Answer to Problem 36E

 The standard deviation is 5.65.

Explanation of Solution

The standard deviation can be obtained as follows:

$\begin{array}{c}\sigma =\sqrt{n\pi \left(1-\pi \right)}\\ =\sqrt{200\left(0.80\right)\left(1-0.80\right)}\\ =\sqrt{200\left(0.80\right)\left(0.20\right)}\\ =\sqrt{32}\\ =5.65\end{array}$

Therefore, the standard deviation is 5.65.

c.

To determine

Find the probability that 150 or more will accept the invitation.

Answer to Problem 36E

The probability that 150 or more will accept the invitation is 0.9679.

Explanation of Solution

The probability that 150 or more will accept the invitation can be obtained as follows:

$\begin{array}{c}P\left(X\ge 150\right)=P\left(X>150-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>149.5\right)\\ =P\left(\frac{X-\mu }{\sigma }>\frac{149.5-160}{5.65}\right)\\ =P\left(Z>\frac{-10.5}{5.65}\right)\\ =P\left(Z>-1.85\right)\\ =1-P\left(Z<-1.85\right)\end{array}$

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 –1.85.
• 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 –1.85 is 0.0321.

Consider,

$\begin{array}{c}P\left(X\ge 150\right)=1-P\left(Z<-1.85\right)\\ =1-0.0321\\ =0.9679\end{array}$

Therefore, the probability that 150 or more will accept the invitation is 0.9679.

d.

To determine

Find the probability that exactly 150 will accept the invitation.

Answer to Problem 36E

The probability that exactly 150 will accept the invitation is 0.0143.

Explanation of Solution

The probability that exactly 150 will accept the invitation can be obtained as follows:

$\begin{array}{c}P\left(X=150\right)=P\left(150-0.5<X<150+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(149.5<X<150.5\right)\\ =P\left(\frac{149.5-160}{5.65}<\frac{X-\mu }{\sigma }<\frac{150.5-160}{5.65}\right)\\ =P\left(\frac{-10.5}{5.65}<Z<\frac{-9.5}{5.65}\right)\\ =P\left(-1.85<Z<-1.68\right)\\ =P\left(Z<-1.68\right)-P\left(Z<-1.85\right)\end{array}$

From the previous Subpart c, the probability of Z less than –1.85 is 0.0321.

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 –1.68.
• 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 –1.68 is 0.0464.

Consider,

$\begin{array}{c}P\left(X=150\right)=P\left(Z<-1.68\right)-P\left(Z<-1.85\right)\\ =0.0464-0.0321\\ =0.0143\end{array}$

Thus, the probability that exactly 150 will accept the invitation is 0.0143.