Chapter 4.5, Problem 13E

A cylindrical hole is drilled in a block, and a cylindrical piston is placed in the hole. The clearance is equal to one-half the difference between the diameters of the hole and the piston. The diameter of the hole is normally distributed with mean 15 cm and standard deviation 0.025 cm, and the diameter of the piston is normally distributed with mean 14.88 cm and standard deviation 0.015 cm.

1. a. Find the mean clearance.
2. b. Find the standard deviation of the clearance.
3. c. What is the probability that the clearance is less than 0.05 cm?
4. d. Find the 25th percentile of the clearance.
5. e. Specifications call for the clearance to be between 0.05 and 0.09 cm. What is the probability that the clearance meets the specification?
6. f. It is possible to adjust the mean hole diameter. To what value should it be adjusted so as to maximize the probability that the clearance will be between 0.05 and 0.09 cm?

To determine

Determine the mean clearance.

Answer to Problem 13E

The mean clearance is 0.06 cm.

Explanation of Solution

Given info:

The diameter of the hole is normally distributed with mean ${\mu }_X=15$ cm and standard deviation ${\sigma }_X=0.025$ cm. The diameter of the piston is normally distributed with mean ${\mu }_Y=14.88$ cm and standard deviation ${\sigma }_Y=0.015$ cm.

Calculation:

The random variable X is defined as the diameter of a hole and the random variable Y as the diameter of a piston. Then X follows Normal with parameters ${\mu }_X=15$ cm and ${\sigma }_X=0.025$ and Y follows Normal with parameters ${\mu }_Y=14.88$ cm and ${\sigma }_Y=0.015$ cm.

The clearance is denoted as C and is defined as $C=0.5X-0.5Y$.

The mean clearance is,

$\begin{array}{c}{\mu }_C=E\left(0.5X-0.5Y\right)\\ =0.5E\left(X\right)-0.5E\left(Y\right)\\ =0.5{\mu }_X-0.5{\mu }_Y\\ =0.5\left(15\right)-0.5\left(14.88\right)\end{array}$

$\begin{array}{c}=7.5-7.44\\ =0.06\end{array}$

Thus, the mean clearance is 0.06 cm.

To determine

Determine the standard deviation of the clearance.

Answer to Problem 13E

The standard deviation of the clearance is 0.01458 cm.

Explanation of Solution

Calculation:

The standard deviation of the clearance is,

$\begin{array}{c}{\sigma }_C=\sqrt{V\left(0.5X-0.5Y\right)}\\ =\sqrt{{\left(0.5\right)}^{2}V\left(X\right)+{\left(0.5\right)}^{2}V\left(Y\right)}\\ =\sqrt{0.25{\sigma }_X^2+0.25{\sigma }_Y^2}\\ =\sqrt{\left(0.25\right){\left(0.025\right)}^2+\left(0.25\right){\left(0.015\right)}^2}\end{array}$

$\begin{array}{c}=\sqrt{\left(0.25\right)\left(6.25\times {10}^{-4}\right)+\left(0.25\right)\left(2.25\times {10}^{-4}\right)}\\ =\sqrt{1.5625\times {10}^{-4}+5.625\times {10}^{-5}}\\ =\sqrt{2.125\times {10}^{-4}}\\ =0.01458\end{array}$

Thus, the standard deviation of the clearance is 0.01458 cm.

To determine

Find the probability that the clearance is less than 0.05.

Answer to Problem 13E

The probability of clearance less than 0.05 is $P\left(C<0.05\right)=0.2451$.

Explanation of Solution

Calculation:

The random variable C is a linear combination of two normal random variables. Then C follows normal with mean ${\mu }_C=0.06$ and standard deviation ${\sigma }_C=0.01458$.

The required probability is,

$P\left(C<0.05\right)$

The formula to convert X values into z score is,

$z=\frac{X-{\mu }_X}{{\sigma }_X}$

Substitute 0.06 for ${\mu }_C$ and 0.01458 for ${\sigma }_C$ in the above formula.

$\begin{array}{c}P\left(C<0.05\right)=P\left(z\le \frac{0.05-0.06}{0.01458}\right)\\ =P\left(z\le \frac{-0.01}{0.01458}\right)\\ =P\left(z\le -0.69\right)\end{array}$

The above probability can be obtained by finding the areas to the left of –0.69.

The shaded region represents the area to the left of –0.69 is shown below:

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

Procedure:

For probability value for $z\le -0.69$,

• Locate –0.6 in the left column of the Table A.2.
• Obtain the value in the corresponding to the row below 0.09.
• The required value is 0.2451.

That is, $P\left(z\le -0.69\right)=0.2451$

Then,

$P\left(C<0.05\right)=0.2451$

Thus, the value of $P\left(C<0.05\right)=0.2451$.

To determine

Find the value of 25^th percentile of the clearance.

Answer to Problem 13E

The value of 25^th percentile of the clearance is 0.0502 cm.

Explanation of Solution

Calculation:

The 25^th percentile is denoted as $p_{25}$.

The 25^th percentile of a random variable C is defined as $p_{25}={\mu }_C+z_{0.25}{\sigma }_C$.

The value of the random variable C for the bottom 0.25 is same as the value of C for the top 0.75. The area to the left of Z is 0.25 and right of Z is 0.75.

The shaded region represents of the area 25^th percentile of a normal random variable is shown below:

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

Procedure:

• Locate an approximate area of 0.2500 in the body of the A.2 table. The area closest to 0.2500 is 0.2514.
• Move left until the first column and note the value as –0.6.
• Move upward until the top row reached and note the value as 0.07.

Thus, the corresponding z-score is –0.67.

Therefore, the 25^th percentile for the random variable C is,

$\begin{array}{c}{p}_{25}={\mu }_C+z_{0.25}{\sigma }_C\\ =0.06+\left(-0.67\right)\left(0.01458\right)\\ =0.06-9.7686\times {10}^{-3}\\ =0.0502\end{array}$

Thus, the 25^th percentile of the clearance is 0.0502 cm.

To determine

Find the probability that the clearance to be between 0.05 cm and 0.09 cm.

Answer to Problem 13E

The probability that the clearance to be between 0.05 cm and 0.09 cm is 0.7352.

Explanation of Solution

Calculation:

The probability that the clearance to be between 0.05 cm and 0.09 cm implies that $P\left(0.05\le C\le 0.09\right)$.

Substitute 0.06 for ${\mu }_C$ and 0.01458 for ${\sigma }_C$ in the above formula.

$\begin{array}{c}P\left(0.05\le C\le 0.09\right)=P\left(\frac{0.05-0.06}{0.01458}\le Z\le \frac{0.09-0.06}{0.01458}\right)\\ =P\left(\frac{-0.01}{0.01458}\le Z\le \frac{0.03}{0.01458}\right)\\ =P\left(-0.69\le Z\le 2.06\right)\end{array}$

The value of $P\left(0.05\le C\le 0.09\right)$ is obtained by finding the difference between the area the left of –0.69 and to the left of 2.06.

The shaded region represents the area between $z=-0.69$ and $z=2.06$ is shown below:

The area under the standard normal curve that lies between $z=-0.69$ and $z=2.06$ is,

$P\left(0.05\le C\le 0.09\right)=P\left(Z\le 2.06\right)-P\left(Z\le -0.69\right)$

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

Procedure:

For z at 2.06,

• Locate 2.0 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.06.

That is, $P\left(Z\le 2.06\right)=0.9803$.

For z at –0.69,

• Locate –0.6 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.06.

That is, $P\left(z\le -0.69\right)=0.2451$

Thus,

$\begin{array}{c}P\left(-0.69\le Z\le 2.06\right)=0.9803-0.2451\\ =0.7352\end{array}$

Thus, the area under the normal curves between $z=-0.69$ and $z=2.06$ is 0.7352.

To determine

Find the value of adjusted mean hole diameter so as to maximize the probability that the clearance to be between 0.05 cm and 0.09 cm.

Answer to Problem 13E

The value of adjusted mean hole diameter has to be adjusted to 15.2cm to maximize the probability that the clearance to be between 0.05 cm and 0.09.

Explanation of Solution

Calculation:

The probability is maximized at the midpoint between 0.05 cm and 0.09 cm.

The midpoint is $\frac{0.05+0.09}{2}=0.07$ cm.

That is, the probability is maximized at ${\mu }_C=0.07$ cm.

The mean diameter of the piston is, ${\mu }_Y=14.88$ cm.

Then, the adjusted mean hole diameter must satisfies the following equation,

${\mu }_C=0.5{\mu }_X-0.5{\mu }_Y$

Substitute ${\mu }_C=0.07$ in the above equation,

$0.07=0.5{\mu }_X-0.5\left(14.88\right)$

On solving,

$\begin{array}{c}{\mu }_X=\frac{0.07+7.44}{0.5}\\ =\frac{7.51}{0.5}\\ =15.02\end{array}$

Thus, the adjusted mean hole diameter is 15.02 cm.

The probability that the clearance to be between 0.05 cm and 0.09 cm implies that $P\left(0.05\le C\le 0.09\right)$.

Substitute 0.07 for ${\mu }_C$ and 0.01458 for ${\sigma }_C$ in the above formula.

$\begin{array}{c}P\left(0.05\le C\le 0.09\right)=P\left(\frac{0.05-0.07}{0.01458}\le Z\le \frac{0.09-0.07}{0.01458}\right)\\ =P\left(\frac{-0.02}{0.01458}\le Z\le \frac{0.02}{0.01458}\right)\\ =P\left(-1.37\le Z\le 1.37\right)\end{array}$

The value of $P\left(0.05\le C\le 0.09\right)$ is obtained by finding the difference between the area the left of –0.69 and to the left of 2.06.

The shaded region represents the area between $z=-1.37$ and $z=1.37$ is shown below:

The area under the standard normal curve that lies between $z=-1.37$ and $z=1.37$ is,

$P\left(0.05\le C\le 0.09\right)=P\left(Z\le 1.37\right)-P\left(Z\le -1.37\right)$

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

Procedure:

For z at 1.37,

• Locate 1.3 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.07.

That is, $P\left(Z\le 1.37\right)=0.9147$.

For z at –1.37,

• Locate –1.3 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.07.

That is, $P\left(z\le -1.37\right)=0.0853$

Thus,

$\begin{array}{c}P\left(-1.37\le Z\le 1.37\right)=0.9147-0.0853\\ =0.8294\end{array}$

Thus, the area under the normal curves between $z=-1.37$ and $z=1.37$ is 0.8294.

Thus, the probability of the clearance to be between 0.05 cm and 0.09 cm is maximized when the mean clearance becomes 0.07.

Therefore, the value of adjusted mean hole diameter has to be adjusted to 15.2cm to maximize the probability that the clearance to be between 0.05 cm and 0.09.