Chapter 6.3, Problem 39P

(a)

To determine

Find the probability that the waiting time would exceed 20 minutes, given that it has exceeded 15 minutes.

Answer to Problem 39P

The probability that the waiting time would exceed 20 minutes, given that it has exceeded 15 minutes is 0.3989.

Explanation of Solution

Calculation:

Z score:

The number of standard deviations the original measurement x is from the value of mean $\mu$ is measured using the z-score or z value. The formula for z score is,

$z=\frac{x-\mu }{\sigma }$

In the formula, x is the raw score, $\mu$ is the mean and $\sigma$ is the standard deviation.

The conditional probability formula is,

$P\left(A|B\right)=\frac{P\left(A\text{ and }B\right)}{P\left(B\right)}$

The variable x denotes the length of time waiting to be seated.

The probability that the waiting time would exceed 20 minutes, given that it has exceeded 15 minutes is,

$\begin{array}{c}P\left(x>20|x>15\right)=\frac{P\left(x>20\text{ and }x>15\right)}{P\left(x>15\right)}\\ =\frac{P\left(x>20\right)}{P\left(x>15\right)}\end{array}$

For $P\left(x>20\right)$:

Substitute x as 20, $\mu$ as 18 and $\sigma$ as 4 in the formula of z score.

$\begin{array}{c}z=\frac{20-18}{4}\\ =\frac{2}{4}\\ =0.5\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 0.5.

• Locate the value 0.5 in column z.
• Locate the value 0.00 in top row.
• The intersecting value of row and column is 0.6915.

The probability is,

$\begin{array}{c}P\left(x>20\right)=P\left(z>0.5\right)\\ =1-P\left(z<0.5\right)\\ =1-0.6915\\ =0.3085\end{array}$

For $P\left(x>15\right)$:

Substitute x as 15, $\mu$ as 18 and $\sigma$ as 4 in the formula of z score.

$\begin{array}{c}z=\frac{15-18}{4}\\ =\frac{-3}{4}\\ =-0.75\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –0.75.

• Locate the value –0.7 in column z.
• Locate the value 0.05 in top row.
• The intersecting value of row and column is 0.2266.

The probability is,

$\begin{array}{c}P\left(x>15\right)=P\left(z>-0.75\right)\\ =1-P\left(z<-0.75\right)\\ =1-0.2266\\ =0.7734\end{array}$

Substituting 0.3085 for $P\left(x>20\right)$ and 0.7734 for $P\left(x>15\right)$ in the required probability:

$\begin{array}{c}P\left(x>20|x>15\right)=\frac{0.3085}{0.7734}\\ =0.3989\end{array}$

Hence, the probability that the waiting time would exceed 20 minutes, given that it has exceeded 15 minutes is 0.3989.

(b)

To determine

Find the probability that the waiting time would exceed 25 minutes, given that it has exceeded 18 minutes.

Answer to Problem 39P

The probability that the waiting time would exceed 25 minutes, given that it has exceeded 18 minutes is 0.0802.

Explanation of Solution

Calculation:

The probability that the waiting time would exceed 25 minutes, given that it has exceeded 18 minutes is,

$\begin{array}{c}P\left(x>25|x>18\right)=\frac{P\left(x>25\text{ and }x>18\right)}{P\left(x>18\right)}\\ =\frac{P\left(x>25\right)}{P\left(x>18\right)}\end{array}$

For $P\left(x>25\right)$:

Substitute x as 25, $\mu$ as 18 and $\sigma$ as 4 in the formula of z score.

$\begin{array}{c}z=\frac{25-18}{4}\\ =\frac{7}{4}\\ =1.75\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 1.75.

• Locate the value 1.7 in column z.
• Locate the value 0.05 in top row.
• The intersecting value of row and column is 0.9599.

The probability is,

$\begin{array}{c}P\left(x>25\right)=P\left(z>1.75\right)\\ =1-P\left(z<1.75\right)\\ =1-0.9599\\ =0.0401\end{array}$

For $P\left(x>18\right)$:

Substitute x as 18, $\mu$ as 18 and $\sigma$ as 4 in the formula of z score.

$\begin{array}{c}z=\frac{18-18}{4}\\ =\frac{0}{4}\\ =0\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 0.

• Locate the value 0.0 in column z.
• Locate the value 0.00 in top row.
• The intersecting value of row and column is 0.5000.

The probability is,

$\begin{array}{c}P\left(x>18\right)=P\left(z>0\right)\\ =1-P\left(z<0\right)\\ =1-0.5\\ =0.5\end{array}$

Substituting 0.0401 for $P\left(x>25\right)$ and 0.5 for $P\left(x>18\right)$ in the required probability:

$\begin{array}{c}P\left(x>25|x>18\right)=\frac{0.0401}{0.5}\\ =0.0802\end{array}$

Hence, the probability that the waiting time would exceed 25 minutes, given that it has exceeded 18 minutes is 0.0802.

(c)

To determine

Show that $P\left(x>20|x>15\right)=\frac{P\left(x>20\right)}{P\left(x>15\right)}$.

Explanation of Solution

Calculation:

Take event A as $x>20$ and event B as $x>15$ in the conditional probability formula.

$P\left(x>20|x>15\right)=\frac{P\left(x>20\text{ and }x>15\right)}{P\left(x>15\right)}$

It is clear that the x is greater than 20 and x is greater than 15 this implies that the value of x is greater than 20. That is,

$P\left(x>20|x>15\right)=\frac{P\left(x>20\right)}{P\left(x>15\right)}$

Hence, it is shown that $P\left(x>20|x>15\right)=\frac{P\left(x>20\right)}{P\left(x>15\right)}$.