Chapter 4.11, Problem 18E

The manufacture of a certain part requires two different machine operations. The time on machine 1 has mean 0.5 hours and standard deviation 0.4 hours. The time on machine 2 has mean 0.6 hours and standard deviation 0.5 hours. The times needed on the machines are independent. Suppose that 100 parts are manufactured.

a.    What is the probability that the total time used by machine 1 is greater than 55 hours?

b.    What is the probability that the total time used by machine 2 is less than 55 hours?

c.    What is the probability that the total time used by both machines together is greater than 115 hours?

d.    What is the probability that the total time used by machine 1 is greater than the total time used by machine 2?

To determine

Find the probability that the total time used by machine 1 is greater than 55 hours.

Answer to Problem 18E

The probability that the total time used by machine 1 is greater than 55 hours is 0.1056.

Explanation of Solution

Given info:

The mean and standard deviation of the time on machine 1 is 0.5 hours and 0.4 hours. The mean and standard deviation of the time on machine 2 is 0.6 hours and 0.5 hours. The time taken by both the machines are independent. The total number of parts manufactured is 100.

Calculation:

The Central Limit Theorem:

The random variables $X_1,\mathrm{...},X_n$ are the random sample from a population with mean $\mu$ and variance ${\sigma }^2$.

The sample mean is $\overline{X}=\frac{X_1+\mathrm{...}+X_n}{n}$ and the sum of the sample observations is $S_n=X_1+\mathrm{...}+X_n$.

Then if n is sufficiently large, $\overline{X}\sim N\left(\mu ,\frac{{\sigma }^2}{n}\right)$ approximately and $S_n\sim N\left(n\mu ,n{\sigma }^2\right)$ approximately.

The random variables $X_1,\mathrm{...},X_{100}$ are defined as the times taken on machine 1 by the 100 parts and the random variables $Y_1,\mathrm{...},Y_{100}$ are defined as the times taken on machine 2 by the 100 parts.

Then for machine 1, the mean is, ${\mu }_{X_i}=0.5$ and standard deviation ${\sigma }_{X_i}=0.4$ and for machine 2, the mean is, ${\mu }_{Y_i}=0.6$ and standard deviation ${\sigma }_{Y_i}=0.5$

The total time on machine 1 is $S_X=X_1+\mathrm{...}+X_{100}$ and total time on machine 2 is $S_Y=Y_1+\mathrm{...}+Y_{100}$.

Then by the central limit theorem $S_X$ is approximately normally distributed.

Mean:

${\mu }_{S_X}=n{\mu }_{X_i}$

Substitute n as 100 and ${\mu }_{X_i}$ as 0.5.

$\begin{array}{c}{\mu }_{S_X}=\left(100\right)\left(0.5\right)\\ =50\end{array}$

Standard deviation:

${\sigma }_{S_X}=\sqrt{n}{\sigma }_{X_i}$

Substitute n as 100 and ${\sigma }_{X_i}$ as 0.4.

$\begin{array}{c}{\sigma }_{S_X}=\sqrt{100}\left(0.4\right)\\ =\left(10\right)\left(0.4\right)\\ =4\end{array}$

Also by the central limit theorem $S_Y$ is approximately normally distributed.

Mean:

${\mu }_{S_Y}=n{\mu }_{Y_i}$

Substitute n as 100 and ${\mu }_{Y_i}$ as 0.6.

$\begin{array}{c}{\mu }_{S_Y}=\left(100\right)\left(0.6\right)\\ =60\end{array}$

Standard deviation:

${\sigma }_{S_Y}=\sqrt{n}{\sigma }_{Y_i}$

Substitute n as 100 and ${\sigma }_{Y_i}$ as 0.5.

$\begin{array}{c}{\sigma }_{S_Y}=\sqrt{100}\left(0.5\right)\\ =\left(10\right)\left(0.5\right)\\ =5\end{array}$

The required probability is, $P\left(S_X>55\right)$.

The formula to convert $S_X$ values into z score is,

$z=\frac{S_X-{\mu }_{S_X}}{{\sigma }_{S_X}}$

Substitute 50 for ${\mu }_{S_X}$ and 4 for ${\sigma }_{S_X}$ in the above formula.

$\begin{array}{c}P\left(S_X>55\right)=1-P\left(S_X\le 55\right)\\ =1-P\left(Z\le \frac{55-50}{4}\right)\\ =1-P\left(Z\le \frac{5}{4}\right)\\ =1-P\left(Z\le 1.25\right)\end{array}$

The above probability can be obtained by finding the areas to the left of 1.25.

The shaded region represents the area to the right of 1.25 is shown below:

Use Table A.2: Cumulative Normal Distribution to find the area.

Procedure:

For z at 1.25,

• Locate 1.2 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.05.

That is, $P\left(Z\le 1.25\right)=0.8944$

Then,

$\begin{array}{c}P\left(S_X>55\right)=1-0.8944\\ =0.1056\end{array}$

Thus, probability that the total time used by machine 1 is greater than 55 hours is 0.1056.

To determine

Find the probability that the total time used by machine 2 is less than 55 hours.

Answer to Problem 18E

The probability that the total time used by machine 2 is less than 55 hours is 0.1587.

Explanation of Solution

Calculation:

The required probability is, $P\left(S_Y<55\right)$.

Substitute 60 for ${\mu }_{S_Y}$ and 5 for ${\sigma }_{S_Y}$ in the formula of z-score.

$\begin{array}{c}P\left(S_Y<55\right)=P\left(S_Y\le 55\right)\\ =P\left(Z\le \frac{55-60}{5}\right)\\ =P\left(Z\le \frac{-5}{5}\right)\\ =P\left(Z\le -1\right)\end{array}$

The above probability can be obtained by finding the areas to the left of –1.

The shaded region represents the area to the left of –1 is shown below:

Use Table A.2: Cumulative Normal Distribution to find the area.

Procedure:

For z at –1,

• Locate –1.0 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.00.

That is, $P\left(Z\le -1\right)=0.1587$

Then,

$P\left(S_Y<55\right)=0.1587$

Thus, the probability that the total time used by machine 2 is less than 55 hours is 0.1587.

To determine

Find the probability that total time used by both machines together is greater than 115 hours.

Answer to Problem 18E

The probability that total time used by both machines together is greater than 115 hours is 0.2177.

Explanation of Solution

Calculation:

Result:

Assume that X and Y are independent random variables with X follows Normal with mean ${\mu }_X$ and standard deviation ${\sigma }_X$ and Y follows Normal with mean ${\mu }_Y$ and standard deviation ${\sigma }_Y$. Then, $X+Y$ follows Normal with mean ${\mu }_X+{\mu }_Y$ and standard deviation $\sqrt{{\sigma }_X^2+{\sigma }_Y^2}$ and $X-Y$ follows Normal with mean ${\mu }_X-{\mu }_Y$ and standard deviation $\sqrt{{\sigma }_X^2+{\sigma }_Y^2}$

Total time used by both machines is denoted as $T=S_X+S_Y$. Both $S_X$ and $S_Y$  are independent. Then by result, the random variable T is approximately normally distributed with mean,

${\mu }_T={\mu }_{S_X}+{\mu }_{S_Y}$

Substitute ${\mu }_{S_X}=50$ and ${\mu }_{S_Y}=60$ in the above equation.

$\begin{array}{c}{\mu }_T=50+60\\ =110\end{array}$

The standard deviation is,

${\sigma }_T=\sqrt{{\sigma }_{S_X}^2+{\sigma }_{S_Y}^2}$

Substitute ${\sigma }_{S_X}=4$ and ${\sigma }_{S_Y}=5$ in the above equation.

$\begin{array}{c}{\sigma }_T=\sqrt{{\left(4\right)}^2+{\left(5\right)}^2}\\ =\sqrt{16+25}\\ =\sqrt{41}\\ =6.403124\end{array}$

The required probability is, $P\left(T>115\right)$.

Substitute 110 for ${\mu }_T$ and 6.403124 for ${\sigma }_T$ in the above formula.

$\begin{array}{c}P\left(T>115\right)=1-P\left(Z\le \frac{115-110}{6.403124}\right)\\ =1-P\left(Z\le \frac{5}{6.403124}\right)\\ =1-P\left(Z\le 0.78\right)\end{array}$

The above probability can be obtained by finding the areas to the left of 0.78.

The shaded region represents the area to the right of 0.78 is shown below:

Use Table A.2: Cumulative Normal Distribution to find the area.

Procedure:

For z at 0.78,

• Locate 0.7 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.08.

That is, $P\left(Z\le 0.78\right)=0.7823$

Then,

$\begin{array}{c}P\left(T>115\right)=1-0.7823\\ =0.2177\end{array}$

Thus, the value of $P\left(T>115\right)=0.2177$.

To determine

Find the probability that total time used by both machine 1 is greater than the total time used by machine 2.

Answer to Problem 18E

The probability that total time used by both machine 1 is greater than the total time used by machine 2 is 0.0594.

Explanation of Solution

Calculation:

The difference between the time on machine 1 and the time on machine2 is denoted as $D=S_X-S_Y$. Both $S_X$ and $S_Y$  are independent. Then by result, the random variable D is approximately normally distributed with mean,

${\mu }_D={\mu }_{S_X}-{\mu }_{S_Y}$

Substitute ${\mu }_{S_X}=50$ and ${\mu }_{S_Y}=60$ in the above equation.

$\begin{array}{c}{\mu }_D=50-60\\ =-10\end{array}$

The standard deviation is,

${\sigma }_D=\sqrt{{\sigma }_{S_X}^2+{\sigma }_{S_Y}^2}$

Substitute ${\sigma }_{S_X}=4$ and ${\sigma }_{S_Y}=5$ in the above equation.

$\begin{array}{c}{\sigma }_D=\sqrt{{\left(4\right)}^2+{\left(5\right)}^2}\\ =\sqrt{16+25}\\ =\sqrt{41}\\ =6.403124\end{array}$

The required probability is, $P\left(D>0\right)$.

Substitute –10 for ${\mu }_D$ and 6.403124 for ${\sigma }_D$ in the above formula.

$\begin{array}{c}P\left(D>0\right)=1-P\left(Z\le \frac{0+10}{6.403124}\right)\\ =1-P\left(Z\le \frac{10}{6.403124}\right)\\ =1-P\left(Z\le 1.56\right)\end{array}$

The above probability can be obtained by finding the areas to the left of 1.56.

The shaded region represents the area to the right of 1.56 is shown below:

Use Table A.2: Cumulative Normal Distribution to find the area.

Procedure:

For z at 1.56,

• Locate 1.5 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.06.

That is, $P\left(Z\le 1.56\right)=0.9406$

Then,

$\begin{array}{c}P\left(D>0\right)=1-0.9406\\ =0.0594\end{array}$

Thus, the value of $P\left(D>0\right)=0.0594$.