Chapter 4.11, Problem 11E

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A shipment is acceptable if at least 440 of the 500 bearings meet the specification. Assume that each shipment contains a random sample of bearings.

a.    What is the probability that a given shipment is acceptable?

b.    What is the probability that more than 285 out of 300 shipments are acceptable?

c.    What proportion of bearings must meet the specification in order that 99% of the shipments are acceptable?

To determine

Find the probability that a given shipment is acceptable.

Answer to Problem 11E

The probability that a given shipment is acceptableis 0.9418.

Explanation of Solution

Given info:

About 90% of the bearings meet a thickness specification. In total there are 500 bearings. If at least 440 of 500 bearings meet specification then the shipment is acceptable.

Calculation:

The non-defective bearings in a shipment is denoted as X. Then X follows Binomial with parameters $n=500$ and $p=0.90$.

The normal distribution can be used to approximate the binomial distribution with,

Mean:

${\mu }_X=np$

Substitute 500 for n and 0.9 for p in the above equation.

$\begin{array}{c}{\mu }_X=\left(500\right)\left(0.9\right)\\ =450\end{array}$

Standard deviation:

${\sigma }_X=\sqrt{np(1-p)}$

Substitute 500 for n and 0.9 for p in the above equation.

$\begin{array}{c}{\sigma }_X=\sqrt{\left(500\right)\left(0.9\right)\left(1-0.9\right)}\\ =\sqrt{\left(500\right)\left(0.9\right)\left(0.1\right)}\\ =\sqrt{45}\\ =6.7082\end{array}$

The approximate binomial probability using the normal distribution is,

$P\left(X\ge a\right)\approx P\left(X>a-0.5\right)$

By using continuity correction, the value 0.5 is subtracted from the value of 440.

Substitute 440 for a,

$\begin{array}{c}P\left(X\ge 440\right)=P\left(X>440-0.5\right)\\ =P\left(X>439.5\right)\end{array}$

The formula to convert X values into z score is,

$z=\frac{X-np}{\sqrt{np(1-p)}}$

Substitute 450for np and6.7082for $\sqrt{np(1-p)}$.

$\begin{array}{c}P\left(X>439.5\right)=P\left(\frac{X-np}{\sqrt{np(1-p)}}>\frac{439.5-450}{6.7082}\right)\\ =P\left(z>\frac{-10.5}{6.7082}\right)\\ =1-P\left(z\le -1.57\right)\end{array}$

The above probability can be obtained by finding the areas to the left of –1.57.

The shaded region represents the area to the right of –1.57 is shown below:

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

Procedure:

For z at –1.57,

• Locate –1.5 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.07.

That is, $P\left(z\le -1.57\right)=0.0582$

Then,

$\begin{array}{c}P\left(X\ge 440\right)=1-0.0582\\ =0.9418\end{array}$

Thus, the probability that a given shipment is acceptable is 0.9418.

To determine

Find probability that more than 285 out of 300 shipments are acceptable.

Answer to Problem 11E

The probability that more than 285 out of 300 shipments are acceptable is 0.2327.

Explanation of Solution

Calculation:

The random variable Y is defined as the number of shipments out of 300 that are acceptable.

From part (a), the probability that a given shipment is acceptable is 0.9418.

Then, the random variable Y follows binomial with $n=300$ and $p=0.9418$.

The normal distribution can be used to approximate the binomial distribution with,

Mean:

${\mu }_Y=np$

Substitute 300 for n and 0.9418 for p in the above equation.

$\begin{array}{c}{\mu }_Y=\left(300\right)\left(0.9418\right)\\ =282.54\end{array}$

Standard deviation:

${\sigma }_X=\sqrt{np(1-p)}$

Substitute n as 300 and 0.9418 for p in the above equation.

$\begin{array}{c}{\sigma }_X=\sqrt{\left(300\right)\left(0.9418\right)\left(1-0.9418\right)}\\ =\sqrt{\left(300\right)\left(0.9418\right)\left(0.0582\right)}\\ =\sqrt{16.444}\\ =4.0551\end{array}$

The approximate binomial probability using the normal distribution is,

$P\left(Y>a\right)\approx P\left(Y\ge a+0.5\right)$

By using continuity correction, the value 0.5 is added from the value of 285.

Substitute 285 for a,

$\begin{array}{c}P\left(Y>285\right)=P\left(Y\ge 285+0.5\right)\\ =P\left(X\ge 285.5\right)\end{array}$

The formula to convert X values into z score is,

$z=\frac{X-np}{\sqrt{np(1-p)}}$

Substitute 282.54for np and4.0551for $\sqrt{np(1-p)}$.

$\begin{array}{c}P\left(Y\ge 285.5\right)=P\left(\frac{Y-np}{\sqrt{np(1-p)}}>\frac{285.5-282.54}{4.0551}\right)\\ =P\left(z>\frac{2.96}{4.0551}\right)\\ =1-P\left(z\le 0.73\right)\end{array}$

The above probability can be obtained by finding the areas to the left of 0.73.

The shaded region represents the area to the right of 0.73 is shown below:

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

Procedure:

For z at 0.73,

• Locate0.7 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.03.

That is, $P\left(z\le 0.73\right)=0.7673$

Then,

$\begin{array}{c}P\left(Y>285.5\right)=1-0.7673\\ =0.2327\end{array}$

Thus, the probability that more than 285 out of 300 shipments are acceptable is 0.2327.

To determine

Find proportion of bearings must meet the specification in order that 99% of the shipments are acceptable.

Answer to Problem 11E

The proportion of bearings must meet the specification in order that 99% of the shipments are acceptable is 0.909.

Explanation of Solution

Calculation:

The required proportion of defective bearings is denoted as p and the number of defective bearings is denoted as X.

Then, the random variable X follows binomial with parameters $n=500$ and $p$.

The normal distribution can be used to approximate the binomial distribution with,

Mean:

${\mu }_X=np$

Substitute 500 for n in the above equation.

${\mu }_X=500p$

Standard deviation:

${\sigma }_X=\sqrt{np(1-p)}$

Substitute n as 500 in the above equation.

${\sigma }_X=\sqrt{500p\left(1-p\right)}$

The probability that a shipment is acceptable is $P\left(X\ge 440\right)=0.99$.

The approximate binomial probability using the normal distribution is,

$P\left(X\ge a\right)\approx P\left(X>a-0.5\right)$

By using continuity correction, the value 0.5 is added from the value of 285.

Substitute 440 for a,

$\begin{array}{c}P\left(X\ge 440\right)=P\left(X>440-0.5\right)\\ =P\left(X>439.5\right)\end{array}$

The formula to convert X values into z score is,

$z=\frac{X-np}{\sqrt{np(1-p)}}$

Substitute 500p for np and $\sqrt{500p\left(1-p\right)}$ for $\sqrt{np(1-p)}$.

$\begin{array}{c}P\left(X\ge 489.5\right)=P\left(\frac{X-np}{\sqrt{np(1-p)}}>\frac{439.5-500p}{\sqrt{500p\left(1-p\right)}}\right)\\ =P\left(Z>\frac{439.5-500p}{\sqrt{500p\left(1-p\right)}}\right)\end{array}$

The value of the random variable X for the top 0.99 is same as the value of X for the top 0.01. The area to the left of X is 0.01 and right of X is 0.99.

The shaded region represents of the area of 99% of the shipments that are acceptable is shown below:

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

Procedure:

• Locate an approximate area of 0.0100 in the body of the A.2 table. The area closest to 0.0100 is 0.0099.
• Move left until the first column and note the value as –2.3.
• Move upward until the top row reached and note the value as 0.03.

Thus, the corresponding z-score is –2.33.

The z-score can be expressed in terms of p as follows:

$-2.33=\frac{\left(439.5-500p\right)}{\sqrt{500p\left(1-p\right)}}$

Taking square on both sides,

$5.4289=\frac{{\left(439.5-500p\right)}^2}{500p\left(1-p\right)}$

Implies,

$2,714.45p-2,714.45p^2=193,160.25+250,000p^2-439,500p$

That is,

$252,714.45p^2-442,214.45p+193,160.25=0$

The formula for solving the quadratic equation of the form $a{p}^2+bp+c=0$ is,

$p=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Substitute $a=252,714.45,b=-442,214.45\text{and }c=193,160.25$ in the above equation.

$\begin{array}{c}p=\frac{442,214.45\pm \sqrt{{\left(-442,214.45\right)}^2-4\left(252,714.45\right)\left(193,160.25\right)}}{2\left(252,714.45\right)}\\ =\frac{442,214.45\pm \sqrt{1.9555\times {10}^{11}-1.9526\times {10}^{11}}}{505,428.9}\\ =\frac{442,214.45\pm \sqrt{290,000,000}}{505,428.9}\\ =\frac{442,214.45\pm 17,029.3864}{505,428.9}\end{array}$

$\begin{array}{c}=\frac{459,243.8364}{505,428.9}\text{or }\frac{425,185.0636}{505,428.9}\\ =0.909\text{ or 0}\text{.841}\end{array}$

On solving this quadratic equation the value of p is,

$p=0.909$

Since 0.841 is a spurious root.

Thus, the proportion of bearings must meet the specification in order that 99% of the shipments are acceptable is 0.909.