Chapter 4.5, Problem 81E

Sales delay is the elapsed time between the manufacture of a product and its sale. According to the article “Warranty Claims Data Analysis Considering Sales Delay” (Quality and Reliability Engr. Intl., 2013:113–123), it is quite common for investigators to model sales delay using a lognormal distribution. For a particular product, the cited article proposes this distribution with parameter values µ = 2.05 and σ² = .06 (here the unit for delay is months).

a. What are the variance and standard deviation of delay time?

b. What is the probability that delay time exceeds12 months?

c. What is the probability that delay time is within one standard deviation of its mean value?

d. What is the median of the delay time distribution?

e. What is the 99th percentile of the delay time distribution?

f. Among 10 randomly selected such items, how many would you expect to have a delay time exceeding8 months?

To determine

Find the variance and standard deviation of the delay time.

Answer to Problem 81E

The variance of the delay time is 3.97.

The standard deviation of the delay time is 1.99 months.

Explanation of Solution

Given info:

The sales delay time of the manufactured product follows lognormal distribution with parameter values $\mu =2.05\text{ and }{\sigma }^2=0.06$.

Calculation:

Variance of the lognormal distribution:

The variance of the lognormal distribution is,

$V\left(X\right)=e^{2\mu +{\sigma }^2}\times \left(e^{{\sigma }^2}-1\right)$

The standard deviation of the lognormal distribution is,

$\begin{array}{c}{\sigma }_X=\sqrt{V\left(X\right)}\\ =\sqrt{e^{2\mu +{\sigma }^2}\times \left(e^{{\sigma }^2}-1\right)}\end{array}$

The variance is obtained as follows:

$V\left(X\right)=e^{2\mu +{\sigma }^2}\times \left(e^{{\sigma }^2}-1\right)$

Substitute $\mu =2.05\text{ and }{\sigma }^2=0.06$

$\begin{array}{c}V\left(X\right)=e^{2\left(2.05\right)+0.06}\times \left(e^{0.06}-1\right)\\ =e^{4.1+0.06}\times e^{0.06}-1\\ =e^{4.16}\times 0.062\\ =64.07\times 0.062\\ =3.97\end{array}$

Thus the variance of the delay time is 3.97.

The standard deviation is obtained as follows:

$\begin{array}{c}{\sigma }_X=\sqrt{3.97}\\ =1.99\end{array}$

Thus, the standard deviation of the delay time is 1.99 months.

To determine

Find the probability that delay time exceeds 12 months.

Answer to Problem 81E

The probability that the delay time exceeds 12 months is 0.0375.

Explanation of Solution

Calculation:

Lognormal distribution:

The cumulative density function of X is,

$\begin{array}{c}P\left(x;\mu ,\sigma \right)=P\left(Z\le \frac{\ln \left(x\right)-\mu }{\sigma }\right)\\ =\Phi \left(\frac{\ln \left(x\right)-\mu }{\sigma }\right)\end{array}$

The probability that delay time exceeds 12 months is obtained as shown below:

$\begin{array}{c}P\left(X>12\right)=1-P\left(X\le 12\right)\\ =1-P\left(Z\le \frac{\ln \left(12\right)-\mu }{\sigma }\right)\\ =1-\Phi \left(\frac{\ln \left(12\right)-\mu }{\sigma }\right)\end{array}$

Substitute $\mu =2.05\text{ and }\sigma =\sqrt{0.06}$,

$\begin{array}{c}P\left(X>12\right)=1-P\left(Z\le \frac{\ln \left(12\right)-2.05}{\sqrt{0.06}}\right)\\ =1-\Phi \left(\frac{\ln \left(12\right)-2.05}{\sqrt{0.06}}\right)\\ =1-\Phi \left(\frac{2.4849-2.05}{0.2449}\right)\\ =1-\Phi \left(\frac{0.4349}{0.2449}\right)\\ =1-\Phi \left(1.78\right)\end{array}$

Use Table A-3: Standard Normal Curve Areas to find the z value.

For $z=1.78$,

Procedure:

• Locate 1.70 in the left column of the table.
• Obtain the value in the corresponding row below 0.08.

That is, $P\left(z\le 1.76\right)=0.9625$

The probability of the event is given below:

$\begin{array}{c}P\left(X>12\right)=1-\Phi \left(1.76\right)\\ =1-0.9625\\ =0.0375\end{array}$

Thus, the probability that delay time exceeds 12 months is 0.0375.

To determine

Find the probability that delay time is within one standard deviation of its mean value.

Answer to Problem 81E

The probability that the delay time is within one standard deviation of its mean value is 0.7015.

Explanation of Solution

Calculation:

The probability that delay time is within one standard deviation of its mean value is obtained as shown below:

$P\left(\left|X-{\mu }_X\right|<{\sigma }_X\right)=P\left({\mu }_X-{\sigma }_X<X<{\mu }_X+{\sigma }_X\right)$

Mean of lognormal distribution:

The mean of lognormal distribution is,

$E\left(X\right)=e^{\mu +\frac{{\sigma }^2}{2}}$

Substitute $\mu =2.05\text{ and }\sigma =\sqrt{0.06}$,

$\begin{array}{c}E\left(X\right)=e^{2.05+\frac{0.06}{2}}\\ =e^{2.05+0.03}\\ =e^{2.08}\\ =8.00\end{array}$

Substitute $\mu =2.05\text{ and }\sigma =\sqrt{0.06}$,

$\begin{array}{c}P\left(\left|X-2.05\right|<\sqrt{0.06}\right)=P\left(8.00-1.99<X<8.00+1.99\right)\\ =P\left(6.01<X<9.99\right)\\ =P\left(X<9.99\right)-P\left(X<6.01\right)\\ =P\left(Z\le \frac{\ln \left(9.99\right)-2.05}{\sqrt{0.06}}\right)-P\left(Z\le \frac{\ln \left(12\right)-2.05}{\sqrt{0.06}}\right)\end{array}$

                                $\begin{array}{c}=\Phi \left(\frac{\ln \left(9.99\right)-2.05}{\sqrt{0.06}}\right)-\Phi \left(\frac{\ln \left(6.01\right)-2.05}{\sqrt{0.06}}\right)\\ =\Phi \left(\frac{2.30-2.05}{\sqrt{0.06}}\right)-\Phi \left(\frac{1.79-2.05}{\sqrt{0.06}}\right)\\ =\Phi \left(\frac{0.25}{0.245}\right)-\Phi \left(\frac{-0.26}{0.245}\right)\\ =\Phi \left(1.02\right)-\Phi \left(-1.06\right)\end{array}$

Use Table A-3: Standard Normal Curve Areas to find the z value.

Procedure:

For $z=1.02$,

• Locate 1.00 in the left column of the table.
• Obtain the value in the corresponding row below 0.02.

That is, $P\left(z\le 1.02\right)=0.8461$

For $z=-1.06$,

• Locate –1.00 in the left column of the table.
• Obtain the value in the corresponding row below 0.06.

That is, $P\left(z\le -1.06\right)=0.1446$

The probability of the event is given below:

$\begin{array}{c}P\left(\left|X-2.05\right|<\sqrt{0.06}\right)=\Phi \left(1.02\right)-\Phi \left(-1.06\right)\\ =0.8461-0.1446\\ =0.7015\end{array}$

Thus, the probability that delay time is within one standard deviation of its mean value is 0.7015.

To determine

Find the median of the delay time.

Answer to Problem 81E

The median of the delay time is 7.77 months.

Explanation of Solution

Calculation:

The median $\tilde{\mu }$ of the normal distribution is obtained as given below:

Substitute $x=\tilde{\mu }$,

Here, $\tilde{\mu }$ means the median of the lognormal distribution.

$\begin{array}{c}F\left(x=\tilde{\mu }\right)=P\left(Z\le \frac{\ln \left(\tilde{\mu }\right)-\mu }{\sigma }\right)\\ 0.5=\Phi \left(\frac{\ln \left(\tilde{\mu }\right)-\mu }{\sigma }\right)\end{array}$

Since $\Phi \left(0\right)=0.5$,

Substitute $\mu =2.05\text{ and }\sigma =\sqrt{0.06}$

$\begin{array}{c}0.5=\Phi \left(\frac{\ln \left(\tilde{\mu }\right)-2.05}{\sqrt{0.06}}\right)\\ \left(\frac{\ln \left(\tilde{\mu }\right)-2.05}{\sqrt{0.06}}\right)={\Phi }^{-1}\left(0.5\right)\end{array}$

Leftmost value of z:

The cumulative area to the left of z is 0.5.

Procedure:

• Locate the nearest value of 0.5 in the body of the Table A-3.
• Move left until the first column and note the value as 0.
• Move upward until the top row is reached and note the value as 0.00.

Thus, ${\Phi }^{-1}\left(0.5\right)=0$

$\begin{array}{c}\left(\frac{\ln \left(\tilde{\mu }\right)-2.05}{\sqrt{0.06}}\right)=0\\ \ln \left(\tilde{\mu }\right)-2.05=0\\ \ln \left(\tilde{\mu }\right)=2.05\\ \tilde{\mu }=e^{2.05}\\ =7.77\end{array}$

Thus, the median of the delay time is 7.77 months.

To determine

Find the 99^th percentile of the delay time of the distribution.

Answer to Problem 81E

The 99^th percentile of the delay time of the distribution is 13.75 months.

Explanation of Solution

Calculation:

Percentiles of lognormal distribution:

$\left(\frac{\ln \left(z_{0.99}\right)-\mu }{\sigma }\right)={\Phi }^{-1}\left(z_{0.99}\right)$

The 99^th percentile of the delay time of the distribution is obtained as shown below:

Use Table A-3: Standard Normal Curve Areas to find the cut of values z.

Rightmost value of z:

The Table A-3: Standard Normal Curve Areas applies only for cumulative areas from the left. Thus,

$\begin{array}{c}\left(\begin{array}{l}\text{Cumulative area to the }\\ \text{left of rightmost value of }z\end{array}\right)=1-\left(\begin{array}{l}\text{Cumulative area to the }\\ \text{right of rightmost value of }z\end{array}\right)\\ =1-0.99\\ =0.01\end{array}$

The cumulative area to the left of z is 0.01.

Procedure:

• Locate the nearest value of 0.01 in the body of the Table A-3.
• Move left until the first column and note the value as 0.00.
• Move upward until the top row is reached and note the value as 0.01.

Substituting the values $\mu =2.05\text{ and }\sigma =\sqrt{0.06}$ ,

$\begin{array}{c}\left(\frac{\ln \left(z_{0.99}\right)-2.05}{\sqrt{0.06}}\right)=2.33\\ \ln \left(z_{0.99}\right)-2.05=0.571\\ \ln \left(z_{0.99}\right)=2.62\\ z_{0.99}=e^{2.62}\\ z_{0.99}=13.75\end{array}$

Thus, the 99^th percentile of the delay time of the distribution is 13.75 months.

To determine

Find the delay time greater than 8 months for 10 selected items.

Answer to Problem 81E

The delay time exceeds 8 months for 10 selected items is 4.522.

Explanation of Solution

Calculation:

The probability that delay time exceeds 8months is obtained as shown below:

$\begin{array}{c}P\left(X>8\right)=1-P\left(X\le 8\right)\\ =1-\Phi \left(\frac{\ln \left(8\right)-\mu }{\sigma }\right)\end{array}$

Substitute $\mu =2.05\text{ and }\sigma =\sqrt{0.06}$,

$\begin{array}{c}P\left(X>8\right)=1-P\left(Z\le \frac{\ln \left(8\right)-2.05}{\sqrt{0.06}}\right)\\ =1-\Phi \left(\frac{\ln \left(8\right)-2.05}{\sqrt{0.06}}\right)\\ =1-\Phi \left(\frac{2.08-2.05}{0.245}\right)\\ =1-\Phi \left(\frac{0.03}{0.245}\right)\\ =1-\Phi \left(0.12\right)\end{array}$

Use Table A-3: Standard Normal Curve Areas to find the z value.

For $z=0.12$,

Procedure:

• Locate 0.10 in the left column of the table.
• Obtain the value in the corresponding row below 0.02.

That is, $P\left(z\le 0.12\right)=0.5478$

The probability of the event is given below:

$\begin{array}{c}P\left(X>8\right)=1-\Phi \left(0.12\right)\\ =1-0.5478\\ =0.4522\end{array}$

Thus, the probability that delay time exceeds 8 months is 0.4522.

The delay time greater than 8 months for the 10 items is obtained as shown below:

$\begin{array}{c}10\times P\left(X>8\right)=10\times 0.4522\\ =4.522\end{array}$

Thus, the delay time exceeds 8 months for the 10 items is 4.522.