Chapter 7, Problem 57CE

a.

To determine

Find the likelihood that 32 or more consider nutrition important.

Answer to Problem 57CE

The likelihood that 32 or more consider nutrition important is 0.9678.

Explanation of Solution

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

• There are only two mutually exclusive outcomes, nutrition is a top priority in their lives and nutrition is not a top priority in their lives.
• The number of trials is fixed, that is 60 men.
• The probability is constant for each trial, which is 0.64.
• The trials are independent of each other.

Thus, the problem satisfies all the conditions of a binomial distribution.

The mean can be obtained as follows:

$\begin{array}{c}\mu =n\pi \\ =60\left(0.64\right)\\ =38.4\end{array}$

The expected number of men who feel nutrition is important is 38.4.

The standard deviation can be obtained as follows:

$\begin{array}{c}\sigma =\sqrt{n\pi \left(1-\pi \right)}\\ =\sqrt{60\left(0.64\right)\left(1-0.64\right)}\\ =\sqrt{60\left(0.64\right)\left(0.36\right)}\\ =\sqrt{13.824}\end{array}$

   $=3.72$   

The standard deviation of men who feel nutrition is important is 3.72.

The conditions for normal approximation to the binomial distribution are checked below:

The number of men $\left(n\right)$ is 60 and probability that nutrition is a top priority in their lives. $\left(\pi \right)$ is 0.64.

Condition 1:

$\begin{array}{c}n\pi =60\left(0.64\right)\\ =38.4\\ >5\end{array}$

The condition 1 is satisfied.

Condition 2:

$\begin{array}{c}n\left(1-\pi \right)=60\left(1-0.64\right)\\ =60\left(0.36\right)\\ =21.6\\ >5\end{array}$

The condition 2 is satisfied.

The conditions 1 and 2 for normal approximation to the binomial distribution are satisfied.

Let the random variable X be the number of number of men who consider nutrition important, which follows normal distribution with population mean $\mu$ as 38.4 men and population standard deviation $\sigma$ as 3.72 men.

The likelihood that 32 or more consider nutrition important can be obtained as follows:

$\begin{array}{c}P\left(X\ge 32\right)=P\left(X\ge 32-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>31.5\right)\\ =1-P\left(\frac{X-\mu }{\sigma }<\frac{31.5-38.4}{3.72}\right)\\ =1-P\left(\frac{X-\mu }{\sigma }<\frac{-6.9}{3.72}\right)\end{array}$

               $=1-P\left(Z<-1.85\right)$

Step-by-step procedure to obtain probability of Z less than –1.85 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 Spreadsheet.

The output obtained using Excel is represented as follows:

From the above output, the probability of Z less than –1.85 is 0.0322.

Now consider the following:

$\begin{array}{c}P\left(X\ge 32\right)=1-P\left(Z<-1.85\right)\\ =1-0.0322\\ =0.9678\end{array}$

Therefore, the likelihood that 32 or more consider nutrition important is 0.9678.

b.

To determine

Find the likelihood that 44 or more consider nutrition important.

Answer to Problem 57CE

The likelihood that 44 or more consider nutrition important is 0.0853.

Explanation of Solution

The likelihood that 44 or more consider nutrition important can be obtained as follows:

$\begin{array}{c}P\left(X\ge 44\right)=P\left(X\ge 44-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>43.5\right)\\ =1-P\left(\frac{X-\mu }{\sigma }<\frac{43.5-38.4}{3.72}\right)\\ =1-P\left(\frac{X-\mu }{\sigma }<\frac{5.1}{3.72}\right)\end{array}$

               $=1-P\left(Z<1.37\right)$

Step-by-step procedure to obtain probability of Z less than 1.37 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 1.37.
• Enter Cumulative as TRUE.
• Click OK, the answer appears in Spreadsheet.

The output obtained using Excel is represented as follows:

From the above output, the probability of Z less than 1.37 is 0.9147.

Now consider the following:

$\begin{array}{c}P\left(X\ge 44\right)=1-P\left(Z<1.37\right)\\ =1-0.9147\\ =0.0853\end{array}$

Therefore, the likelihood that 44 or more consider nutrition important is 0.0853.

c.

To determine

Find the probability that more than 32 but fewer than 43 consider nutrition important.

Answer to Problem 57CE

The probability that more than 32 but fewer than 43 consider nutrition important is 0.8084.

Explanation of Solution

The probability that more than 32 but fewer than 43 consider nutrition important can be obtained as follows:

$\begin{array}{c}P\left(32<X<43\right)=P\left(32+0.5<X\le 43-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(32.5<X<42.5\right)\\ \text{=P}\left(X<42.5\right)-P\left(X<32.5\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{42.5-38.4}{3.72}\right)-P\left(\frac{X-\mu }{\sigma }<\frac{32-38.4}{3.72}\right)\end{array}$

                        $=P\left(Z<1.10\right)-P\left(Z<-1.59\right)$

Step-by-step procedure to obtain probability of Z less than 1.10 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 1.10.
• Enter Cumulative as TRUE.
• Click OK, the answer appears in Spreadsheet.

The output obtained using Excel is represented as follows:

From the above output, the probability of Z less than 1.10 is 0.8643.

Step-by-step procedure to obtain probability of Z less than –1.59 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 –1.59.
• Enter Cumulative as TRUE.
• Click OK, the answer appears in Spreadsheet.

The output obtained using Excel is represented as follows:

From the above output, the probability of Z less than –1.59 is 0.0559.

Now consider,

$\begin{array}{c}P\left(32<X<43\right)=P\left(Z<1.10\right)-P\left(Z<-1.59\right)\\ =0.8643-0.0559\\ =0.8084\end{array}$

Therefore, the probability that more than 32 but fewer than 43 consider nutrition important is 0.8084.

d.

To determine

Find the probability that exactly 44 consider diet important.

Answer to Problem 57CE

The probability that exactly 44 consider diet important is 0.0348.

Explanation of Solution

The probability that exactly 44 consider diet important can be obtained as follows:

$\begin{array}{c}P\left(X=44\right)=P\left(44-0.5\le X\le 44+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(43.5<X<44.5\right)\\ \text{=P}\left(X<44.5\right)-P\left(X<43.5\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{44.5-38.4}{3.72}\right)-P\left(\frac{X-\mu }{\sigma }<\frac{43.5-38.4}{3.72}\right)\end{array}$

               $=P\left(Z<1.64\right)-P\left(Z<1.37\right)$

Step-by-step procedure to obtain probability of Z less than 1.64 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 1.64.
• Enter Cumulative as TRUE.
• Click OK, the answer appears in Spreadsheet.

The output obtained using Excel is represented as follows:

From the above output, the probability of Z less than 1.64 is 0.9495.

Step-by-step procedure to obtain probability of Z less than 1.37 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 1.37
• Enter Cumulative as TRUE.
• Click OK, the answer appears in Spreadsheet.

The output obtained using Excel is represented as follows:

From the above output, the probability of Z less than 1.37 is 0.9147.

Now consider,

$\begin{array}{c}P\left(X=44\right)=P\left(Z<1.64\right)-P\left(Z<1.37\right)\\ =0.9495-0.9147\\ =0.0348\end{array}$

Therefore, the probability that exactly 44 consider diet important is 0.0348.