Chapter 7, Problem 35E

a.

To determine

Explain if the problem meets the requirements of a binomial problem.

Answer to Problem 35E

Yes, the problem qualifies as a binomial problem.

Explanation of Solution

In order to qualify as a binomial problem it must satisfy the following conditions:

• The number of trials is fixed that is 500 new members.
• There are only two mutually exclusive outcomes, overweight and not overweight.
• The probability is constant for each trial that is 0.30.
• The trials are independent to each other.

Since the problem satisfies all the conditions of a binomial distribution.

Hence, the problem is qualified as a binomial problem.

b.

To determine

Find the probability that 175 or more of the new members are 15 pounds overweight.

Answer to Problem 35E

The probability that 175 or more of the new members are 15 pounds overweight is 0.0085.

Explanation of Solution

It is given that 30% of its new members are 15 pounds overweight and the total number of new members in a metropolitan area is 500.

That is, $n=500$ and $\pi =0.30$.

The mean can be obtained as follows:

$\begin{array}{c}\mu =n\pi \\ =500\left(0.30\right)\\ =150\end{array}$

The standard deviation can be obtained as follows:

$\begin{array}{c}\sigma =\sqrt{n\pi \left(1-\pi \right)}\\ =\sqrt{500\left(0.30\right)\left(1-0.30\right)}\\ =\sqrt{500\left(0.30\right)\left(0.70\right)}\\ =\sqrt{105}\\ =10.25\end{array}$

The probability that 175 or more of the new members are 15 pounds overweight can be obtained as follows:

$\begin{array}{c}P\left(X\ge 175\right)=P\left(X>175-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>174.5\right)\\ =P\left(\frac{X-\mu }{\sigma }>\frac{174.5-150}{10.25}\right)\\ =P\left(Z>\frac{24.5}{10.25}\right)\\ =P\left(Z>2.39\right)\\ =1-P\left(Z<2.39\right)\end{array}$

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 2.39.
• 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 2.39 is 0.9915.

Now consider,

$\begin{array}{c}P\left(X\ge 175\right)=1-P\left(Z<2.39\right)\\ =1-0.9915\\ =0.0085\end{array}$

Therefore, the probability that 175 or more of the new members are 15 pounds overweight is 0.0085.

c.

To determine

Find the probability that 140 or more of the new members are 15 pounds overweight.

Answer to Problem 35E

The probability that 140 or more of the new members are 15 pounds overweight is 0.8462.

Explanation of Solution

The probability that 140 or more of the new members are 15 pounds overweight can be obtained as follows:

$\begin{array}{c}P\left(X\ge 140\right)=P\left(X>140-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>139.5\right)\\ =P\left(\frac{X-\mu }{\sigma }>\frac{139.5-150}{10.25}\right)\\ =P\left(Z>\frac{-10.5}{10.25}\right)\\ =P\left(Z>-1.02\right)\\ =1-P\left(Z<-1.02\right)\end{array}$

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 –1.02.
• 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 –1.02 is 0.1538.

Consider,

$\begin{array}{c}P\left(X\ge 140\right)=1-P\left(Z<-1.02\right)\\ =1-0.1538\\ =0.8462\end{array}$

Therefore, the probability that 140 or more of the new members are 15 pounds overweight is 0.8462.