Chapter 7, Problem 58CE

a.

To determine

Find the expected number of students who fail in the exam.

Find the standard deviation of number of students who fail in the exam.

Answer to Problem 58CE

The expected number of students who fail in the exam is 6.

The standard deviation of number of students who fail in the exam is 2.32

Explanation of Solution

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

• There are only two mutually exclusive outcomes, students fail in the exam and students do not fail in the exam.
• The number of trials is fixed, that is, 60 students.
• The probability is constant for each trial, which is 0.10.
• The trials are independent to each other.

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

The mean can be obtained as follows:

$\begin{array}{c}\mu =n\pi \\ =60\left(0.10\right)\\ =6\end{array}$

Therefore, the expected number of students who fail in the exam is 6.

The standard deviation can be obtained as follows:

$\begin{array}{c}\sigma =\sqrt{n\pi \left(1-\pi \right)}\\ =\sqrt{60\left(0.10\right)\left(1-0.10\right)}\\ =\sqrt{60\left(0.10\right)\left(0.90\right)}\\ =\sqrt{5.4}\end{array}$

     $=2.32$

Therefore, the standard deviation of number of students who fail in the exam is 2.32.

b.

To determine

Find the probability that exactly two students will fail.

Answer to Problem 58CE

The probability that exactly two students will fail is 0.0393.

Explanation of Solution

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

The number of students $\left(n\right)$ is 60 and probability that student fail in the exam $\left(\pi \right)$ is 0.10.

Condition 1:

$\begin{array}{c}n\pi =60\left(0.10\right)\\ =6\\ >5\end{array}$

Condition 1 is satisfied.

Condition 2:

$\begin{array}{c}n\left(1-\pi \right)=60\left(1-0.10\right)\\ =60\left(0.90\right)\\ =54\\ >5\end{array}$

Condition 2 is satisfied.

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

Let the random variable X be the number of students who fail in the exam follows normal distribution with population mean $\mu$ as 6 students and population standard deviation $\sigma$ as 2.32 students.

The probability that exactly two students will fail can be obtained as follows:

$\begin{array}{c}P\left(X=2\right)=P\left(2-0.5\le X\le 2+0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(1.5<X<2.5\right)\\ =P\left(X<2.5\right)-P\left(X<1.5\right)\\ =P\left(\frac{X-\mu }{\sigma }<\frac{2.5-6}{2.32}\right)-P\left(\frac{X-\mu }{\sigma }<\frac{1.5-6}{2.32}\right)\end{array}$

                $=P\left(Z<-1.51\right)-P\left(Z<-1.94\right)$

Step-by-step procedure to obtain the probability of Z less than –1.51 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 –1.51.
• 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 –1.51 is 0.0655.

Step-by-step procedure to obtain the probability of Z less than –1.94 using Excel:

• Click on the Formulas tab in the top menu.
• Select Insert function. Then from the category box, select Statistical and below that NORM.S.DIST.
• Click OK.
• In the dialog box, Enter the Z value as –1.94.
• Enter Cumulative as TRUE.
• Click OK, the answer appears in the 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:

$\begin{array}{c}P\left(X=2\right)=P\left(Z<-1.51\right)-P\left(Z<-1.94\right)\\ =0.0655-0.0262\\ =0.0393\end{array}$

Therefore, the probability that exactly two students will fail is 0.0393.

c.

To determine

Find the probability that at least two students will fail.

Answer to Problem 58CE

The probability that at least two students will fail is 0.9738.

Explanation of Solution

The probability that at least two students will fail can be obtained as follows:

$\begin{array}{c}P\left(X\ge 2\right)=P\left(X\ge 2-0.5\right)\left[\begin{array}{l}\text{Apply the continuity }\\ \text{correction factor}\end{array}\right]\\ =P\left(X>1.5\right)\\ =1-P\left(X<1.5\right)\\ =1-P\left(\frac{X-\mu }{\sigma }<\frac{1.5-6}{2.32}\right)\end{array}$

               $=1-P\left(Z<-1.94\right)$

From Part (b), the probability of Z less than –1.94 is 0.0262.

Now consider,

$\begin{array}{c}P\left(X\ge 2\right)=1-P\left(Z<-1.94\right)\\ =1-0.0262\\ =0.9738\end{array}$

Therefore, the probability that at least two students will fail is 0.9738.