Chapter 7.8, Problem 117E

Suppose that 25% of the fire alarms in a large city are false alarms. Let x denote the number of false alarms in a random sample of 100 alarms. Approximate the following probabilities:

1. a. P(20 ≤ x ≤ 30)
2. b. P(20 < x < 30)
3. c. P(x ≥ 35)
4. d. The probability that x is farther than 2 standard deviations from its mean value

To determine

Find the approximate probability, $P\left(20\le x\le 30\right)$.

Answer to Problem 117E

The approximate probability, $P\left(20\le x\le 30\right)$ is 0.7960.

Explanation of Solution

Calculation:

It is given that the 25% of the fire alarms in a large city are false alarms. Also, 100 alarms are randomly selected.

Define the random variable x number of false alarms. Here, x follows binomial distribution with $n=100\text{ and }p=0.25$

The mean and standard deviation of a binomial distribution is obtained as given below:

$\begin{array}{c}\mu =np\\ =\left(100\right)\left(0.25\right)\\ =25\\ \sigma =\sqrt{np\left(1-p\right)}\\ =\sqrt{\left(100\right)\left(0.25\right)\left(1-0.25\right)}\\ =\sqrt{\left(100\right)\left(0.25\right)\left(0.75\right)}\\ =\sqrt{\text{18}\text{.75}}\\ =4.33\end{array}$

Conditions for binomial to follow normal distribution (approximately):

• $np\ge 10$
• $n\left(1-p\right)\ge 10$

Substitute $n=100\text{ and }p=0.25$

$\begin{array}{c}np=\left(100\right)\left(0.25\right)\\ =25\\ >10\end{array}$

$\begin{array}{c}n\left(1-p\right)=\left(100\right)\left(1-0.25\right)\\ =\left(100\right)\left(0.75\right)\\ =\text{75}\\ >10\end{array}$

Thus, both the conditions are satisfied. Hence, the distribution of x is approximately normal.

The desired probability is approximately the area under the normal curve between 19.5 and 30.5.

The required probability is obtained as given below:

$\begin{array}{c}P\left(20\le x\le 30\right)\approx P\left(19.5\le x\le 30.5\right)\\ =P\left(\frac{19.5-25}{4.33}\le z\le \frac{30.5-25}{4.33}\right)\\ =P\left(\frac{-5.5}{4.33}\le z\le \frac{5.5}{4.33}\right)\\ =P\left(-1.27\le z\le 1.27\right)\\ =P\left(z\le 1.27\right)-P\left(z\le -1.27\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -1.27\right)$.

Procedure:

• In the $z*$ row locate −1.2
• In column locate .07
• The intersection of the row −1.2 and column .07 gives 0.1020

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le 1.27\right)$.

Procedure:

• In the $z*$ row locate 1.2
• In column locate .07
• The intersection of the row 1.2 and column .07 gives 0.8980

The approximated probability is obtained as given below:

$\begin{array}{c}P\left(z\le 1.27\right)-P\left(z\le -1.27\right)=0.8980-0.1020\\ =\text{0}\text{.796}0\end{array}$

Thus, the approximate probability, $P\left(20\le x\le 30\right)$ is 0.7960.

To determine

Find the approximate probability, $P\left(20<x<30\right)$.

Answer to Problem 117E

The approximate probability, $P\left(20<x<30\right)$ is 0.7016.

Explanation of Solution

Calculation:

The desired probability is approximately the area under the normal curve between 20.5 and 29.5.

The required probability is obtained as given below:

$\begin{array}{c}P\left(20<x<30\right)\approx P\left(20.5\le x\le 29.5\right)\\ =P\left(\frac{20.5-25}{4.33}\le z\le \frac{29.5-25}{4.33}\right)\\ =P\left(\frac{-4.5}{4.33}\le z\le \frac{4.5}{4.33}\right)\\ =P\left(-1.04\le z\le 1.04\right)\\ =P\left(z\le 1.04\right)-P\left(z\le -1.04\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -1.04\right)$.

Procedure:

• In the $z*$ row locate −1.0
• In column locate .04
• The intersection of the row −1.0 and column .04 gives 0.1492

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le 1.04\right)$.

Procedure:

• In the $z*$ row locate 1.0
• In column locate .04
• The intersection of the row 1.0 and column .04 gives 0.8508

The approximated probability is obtained as given below:

$\begin{array}{c}P\left(z\le 1.04\right)-P\left(z\le -1.04\right)=0.8508-0.1492\\ =\text{0}\text{.7016}\end{array}$

Thus, the approximate probability, $P\left(20<x<30\right)$ is 0.7016.

To determine

Find the approximate probability, $P\left(x\ge 35\right)$

Answer to Problem 117E

The approximate probability, $P\left(x\ge 35\right)$ is 0.0143.

Explanation of Solution

Calculation:

The desired probability is approximately the area under the normal curve greater than or equal to 34.5.

The required probability is obtained as given below:

$\begin{array}{c}P\left(x\ge 35\right)\approx P\left(x\ge 34.5\right)\\ =P\left(z\ge \frac{34.5-25}{4.33}\right)\\ =P\left(z\ge \frac{9.5}{4.33}\right)\\ =P\left(z\ge 2.19\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le 2.19\right)$.

Procedure:

• In the $z*$ row locate 2.1
• In column locate .09
• The intersection of the row 2.1 and column .09 gives 0.9857

The approximated probability is obtained as given below:

$\begin{array}{c}P\left(z\ge 2.19\right)=1-P\left(z\le 2.19\right)\\ =1-0.9857\\ =\text{0}\text{.0143}\end{array}$

Thus, the approximate probability, $P\left(x\ge 35\right)$ is 0.0143.

To determine

Find the approximate probability that x is farther than 2 standard deviations from its mean value.

Answer to Problem 117E

The approximate probability that x is farther than 2 standard deviations from its mean value is 0.05.

Explanation of Solution

Calculation:

Two standard deviations away from the mean is obtained as given below:

$\begin{array}{c}\mu -2\sigma =25-2\left(4.33\right)\\ =25-\text{8}\text{.66}\\ =\text{16}\text{.34}\\ \mu +2\sigma =25+2\left(4.33\right)\\ =25+\text{8}\text{.66}\\ =\text{33}\text{.66}\end{array}$

Thus, two standard deviations less than the mean is obtained as given below:

$\begin{array}{c}P\left(16.34\le x\le 33.66\right)=P\left(17\le x\le 33\right)\\ \approx P\left(16.5\le x\le 33.5\right)\\ =P\left(\frac{16.5-25}{4.33}\le z\le \frac{33.5-25}{4.33}\right)\\ =P\left(\frac{-8.5}{4.33}\le z\le \frac{8.5}{4.33}\right)\\ =P\left(-1.96\le z\le 1.96\right)\\ =P\left(z\le 1.96\right)-P\left(z\le -1.96\right)\end{array}$

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le -1.96\right)$.

Procedure:

• In the $z*$ row locate −1.9
• In column locate .06
• The intersection of the row −1.9 and column .06 gives 0.0250

Use Table 2 Standard Normal Probabilities (Cumulative z Curve Areas) to obtain $P\left(z\le 1.96\right)$.

Procedure:

• In the $z*$ row locate 1.9
• In column locate .06
• The intersection of the row 1.9 and column .06 gives 0.9750

The approximated probability is obtained as given below:

$\begin{array}{c}P\left(z\le 1.96\right)-P\left(z\le -1.96\right)=0.9750-0.0250\\ =\text{0}\text{.95}\end{array}$

The required probability is obtained is obtained as given below:

$\begin{array}{c}P\left(x\text{ more than 2 standard deviations}\right)=1-\left[P\left(z\le 1.96\right)-P\left(z\le -1.96\right)\right]\\ =1-\text{0}\text{.95}\\ =0.05\end{array}$

Thus, the approximate probability that x is farther than 2 standard deviations from its mean value is 0.05.