Chapter 7, Problem 56CE

a.

To determine

Find the probability that fewer than five employees steal.

Answer to Problem 56CE

The probability that fewer than five employees steal is 0.0262.

Explanation of Solution

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

• There are only two mutually exclusive outcomes, employees steal from their company and employees do not steal from their company.
• The number of trials is fixed, that is 50 people.
• The probability is constant for each trial, which is 0.20.
• 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:

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

The expected number of employees who steal is 10.

The standard deviation can be obtained as follows:

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

   $=2.83$   

The standard deviation of number of employees who steal is 2.83.

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

The number of employees $\left(n\right)$ is 50 and probability that employees steal $\left(\pi \right)$ is 0.20.

Condition 1:

$\begin{array}{c}n\pi =50\left(0.20\right)\\ =10\\ >5\end{array}$

The condition 1 is satisfied.

Condition 2:

$\begin{array}{c}n\left(1-\pi \right)=50\left(1-0.20\right)\\ =50\left(0.80\right)\\ =40\\ >5\end{array}$

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 employees who steal from their company follows normal distribution with population mean $\mu$ as 10 employees and population standard deviation $\sigma$ as 2.83 employees.

The probability that fewer than five employees steal can be obtained as follows:

$\begin{array}{c}P\left(X<5\right)=P\left(X<5-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X<4.5\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{4.5-10}{2.83}\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{-5.5}{2.83}\right)\end{array}$

               $=P\left(Z<-1.94\right)$

Step-by-step procedure to obtain probability of Z less than –1.94 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 –1.94.
• 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 –1.94 is 0.0262.

Now consider the following:

$\begin{array}{c}P\left(X<5\right)=P\left(Z<-1.94\right)\\ =0.0262\end{array}$

Therefore, the probability that fewer than five employees steal is 0.0262.

b.

To determine

Find the probability that more than five employees steal.

Answer to Problem 56CE

The probability that more than five employees steal is 0.9441.

Explanation of Solution

The population mean $\mu$ is 10 employees and population standard deviation $\sigma$ is 2.83 employees.

The probability that more than five employees steal can be obtained as follows:

$\begin{array}{c}P\left(X>5\right)=P\left(X>5+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>5.5\right)\\ =1-P\left(\frac{X-\mu }{\sigma }<\frac{5.5-10}{2.83}\right)\\ =1-P\left(\frac{X-\mu }{\sigma }<\frac{-4.5}{2.83}\right)\end{array}$

               $=1-P\left(Z<-1.59\right)$

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 the following:

$\begin{array}{c}P\left(X>5\right)=1-P\left(Z<-1.59\right)\\ =1-0.559\\ =0.9441\end{array}$

Therefore, the probability that more than five employees steal is 0.9441.

c.

To determine

Find the probability that exactly five employees steal.

Answer to Problem 56CE

The probability that exactly five employees steal is 0.0297.

Explanation of Solution

The probability that exactly five employees steal can be obtained as follows:

$\begin{array}{c}P\left(X=5\right)=P\left(5-0.5\le X\le 5+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(4.5<X<5.5\right)\\ \text{=P}\left(X<5.5\right)-P\left(X<4.5\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{5.5-10}{2.83}\right)-P\left(\frac{X-\mu }{\sigma }<\frac{4.5-10}{2.83}\right)\end{array}$

               $=P\left(Z<-1.59\right)-P\left(Z<-1.94\right)$

From the part (a), the probability of Z less than –1.94 is 0.0262.

From the part (b), the probability of Z less than –1.59 is 0.0559.

Now consider,

$\begin{array}{c}P\left(X=5\right)=P\left(Z<-1.59\right)-P\left(Z<-1.94\right)\\ =0.0559-0.0262\\ =0.0297\end{array}$

Therefore, the probability that exactly five employees steal is 0.0297.

d.

To determine

Find the probability that more than 5 but fewer than 15 employees steal.

Answer to Problem 56CE

The probability that more than 5 but fewer than 15 employees steal is 0.8882.

Explanation of Solution

The probability that more than 5 but fewer than 15 employees steal can be obtained as follows:

$\begin{array}{c}P\left(5<X<15\right)=P\left(5+0.5<X\le 15-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(5.5<X<14.5\right)\\ \text{=P}\left(X<14.5\right)-P\left(X<5.5\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{14.5-10}{2.83}\right)-P\left(\frac{X-\mu }{\sigma }<\frac{5.5-10}{2.83}\right)\end{array}$

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

From the part (b), the probability of Z less than –1.59 is 0.0559.

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 0.9441.

Now consider,

$\begin{array}{c}P\left(5<X<15\right)=P\left(Z<1.59\right)-P\left(Z<-1.59\right)\\ =0.9441-0.0559\\ =0.8882\end{array}$

Therefore, the probability that more than 5 but fewer than 15 employees steal is 0.8882.