Chapter 3.4, Problem 58E

A very large batch of components has arrived at a distributor. The batch can be characterized as accept-able only if the proportion of defective components is at most .10. The distributor decides to randomly select 10 components and to accept the batch only if the number of defective components in the sample is at most 2.

1. a. What is the probability that the batch will be accepted when the actual proportion of defectives is .01? .05? .10? .20? .25?
2. b. Let p denote the actual proportion of defectives in the batch. A graph of P(batch is accepted) as a function of p, with p on the horizontal axis and P(batch is accepted) on the vertical axis, is called the operating characteristic curve for the acceptance sampling plan. Use the results of part (a) to sketch this curve for 0 ≤ p ≤ 1.
3. c. Repeat parts (a) and (b) with “1” replacing “2” in the acceptance sampling plan.
4. d. Repeat parts (a) and (b) with “15” replacing “10” in the acceptance sampling plan.
5. e. Which of the three sampling plans, that of part (a), (c), or (d), appears most satisfactory, and why?

To determine

Find the probability that the batch will be accepted when the actual proportion of defectives is 0.01, 0.05, 0.10, 0.20 and 0.25.

Answer to Problem 58E

The probability that the batch will be accepted for different values of actual proportion of defectives, are,

p	$P\left(X\le 2\right)$
0.01	0.998
0.05	0.9885
0.1	0.9298
0.20	0.6778
0.25	0.5256

Explanation of Solution

Given info:

A distributor decided to select 10 components randomly from a batch. The batch will be accepted if the number of defective components in the sample is at most 2. A batch is characterized as acceptable if the proportion of defective components is at most 0.10.

Calculation:

Let p be the actual proportion of defective in the batch.

Let X be the number of defectives in the sample.

The sample size is 10. The samples are selected randomly and the outcomes are defective or non-defective.

Hence, $X\sim Bin\left(10,p\right)$

It is given that the batch will be accepted if the number of defective components in the sample is at most 2 means the batch will be accepted if $X\le 2$ .

The probability that the batch will be accepted when the actual proportion of defectives is p, is $P\left(X\le 2\right)={\displaystyle \sum _{x=0}^2\left(\begin{array}{l}10\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}}$ .

The probability values:

For p = 0.01:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.01.
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

For p = 0.05:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.05
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

For p = 0.10:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.10
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

For p = 0.2:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.20
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

For p = 0.25:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.25
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

Hence, the probability that the batch will be accepted for different values of actual proportion of defectives and for $c=2$ , are,

p	$P\left(X\le 2\right)$
0.01	0.998
0.05	0.9885
0.1	0.9298
0.20	0.6778
0.25	0.5256

To determine

Sketch the OC curve for the accepting sampling plan.

Answer to Problem 58E

The graph is given below:

Explanation of Solution

Calculation:

From part (a) it is found that, the probability that the batch will be accepted for different values of actual proportion of defectives, are,

p	$f\left(p\right)=P\left(X\le 2\right)$
0.01	0.998
0.05	0.9885
0.1	0.9298
0.20	0.6778
0.25	0.5256

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Graph > Scatterplot.
• Choose With Connect Line, and then click OK.
• Under Y variables, enter a column of f(p).
• Under X variables, enter a column of p.
• Click OK.

Output using MINITAB software is given below:

To determine

Find the probability that the batch will be accepted when the actual proportion of defectives is 0.01, 0.05, 0.10, 0.20 and 0.25 with “1” replacing “2”.

Sketch the OC curve for the accepting sampling plan.

Answer to Problem 58E

The probability that the batch will be accepted when the actual proportion of defectives is 0.01, 0.05, 0.10, 0.20 and 0.25 are 0.9957, 0.9138, 0.7360, 0.3758, and 0.2440 respectively.

The OC curve is given below

Explanation of Solution

Given info:

The c value 2 is replaced by 1.

Calculation:

Let p be the actual proportion of defective in the batch.

Let X be the number of defectives in the sample.

The sample size is 10. The samples are selected randomly and the outcomes are defective or non-defective.

Hence, $X\sim Bin\left(10,p\right)$

It is given that the batch will be accepted if the number of defective components in the sample is at most 1 means the batch will be accepted if $X\le 1$ .

The probability that the batch will be accepted when the actual proportion of defectives is p, is $P\left(X\le 1\right)={\displaystyle \sum _{x=0}^1\left(\begin{array}{l}10\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}}$ .

The probability values:

For p = 0.01:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.01
• Choose Cumulative probability.
• In Input constant, enter 1.
• Click OK.

Output using MINITAB software is given below:

For p = 0.05:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.05
• Choose Cumulative probability.
• In Input constant, enter 1.
• Click OK.

Output using MINITAB software is given below:

For p = 0.10:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.10
• Choose Cumulative probability.
• In Input constant, enter 1.
• Click OK.

Output using MINITAB software is given below:

For p = 0.2:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.20
• Choose Cumulative probability.
• In Input constant, enter 1.
• Click OK.

Output using MINITAB software is given below:

For p = 0.25:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 10 and event probability as 0.25.
• Choose Cumulative probability.
• In Input constant, enter 1.
• Click OK.

Output using MINITAB software is given below:

Thus, the probability that the batch will be accepted for different values of actual proportion of defectives, $n=10$  and for $c=1$ , are,

p	$P\left(X\le 2\right)$
0.01	0.9957
0.05	0.9138
0.1	0.7360
0.20	0.3758
0.25	0.2440

OC curve:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Graph > Scatterplot.
• Choose With Connect Line, and then click OK.
• Under Y variables, enter a column of f(p).
• Under X variables, enter a column of p.
• Click OK.

Output using MINITAB software is given below:

To determine

Find the probability that the batch will be accepted when the actual proportion of defectives is 0.01, 0.05, 0.10, 0.20 and 0.25 with “15” replacing “10”.

Sketch the OC curve for the accepting sampling plan.

Answer to Problem 58E

The probability that the batch will be accepted for different values of actual proportion of defectives, $n=15$  and for $c=2$ , are,

p	$P\left(X\le 2\right)$
0.01	0.996
0.05	0.9638
0.1	0.8159
0.20	0.3980
0.25	0.2360

The OC curve is given below

Explanation of Solution

Given info:

10 is replaced by 15.

Calculation:

Let p be the actual proportion of defective in the batch.

Let X be the number of defectives in the sample.

The sample size is 15. The samples are selected randomly and the outcomes are defective or non-defective.

Hence, $X\sim Bin\left(15,p\right)$

It is given that the batch will be accepted if the number of defective components in the sample is at most 2 means the batch will be accepted if $X\le 2$ .

The probability that the batch will be accepted when the actual proportion of defectives is p, is $P\left(X\le 2\right)={\displaystyle \sum _{x=0}^2\left(\begin{array}{l}15\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x}}$ .

The probability values:

For p = 0.01:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 15 and event probability as 0.01
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

For p = 0.05:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 15 and event probability as 0.05
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

For p = 0.10:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 15 and event probability as 0.10
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

For p = 0.2:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 15 and event probability as 0.20
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

For p = 0.25:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Calc > Probability Distribution> Binomial Distribution.
• Enter number of trials as 15 and event probability as 0.25
• Choose Cumulative probability.
• In Input constant, enter 2.
• Click OK.

Output using MINITAB software is given below:

Hence, the probability that the batch will be accepted for different values of actual proportion of defectives, $n=15$  and for $c=2$ , are,

p	$P\left(X\le 2\right)$
0.01	0.996
0.05	0.9638
0.1	0.8159
0.20	0.3980
0.25	0.2360

OC curve:

Software procedure:

Step by step procedure to obtain the probability values using the MINITAB software:

• Choose Graph > Scatterplot.
• Choose With Connect Line, and then click OK.
• Under Y variables, enter a column of f(p).
• Under X variables, enter a column of p.
• Click OK.

Output using MINITAB software is given below:

To determine

Find the most satisfactory sampling plan among a, c and d.

Answer to Problem 58E

Plan in part (d) is most satisfactory.

Explanation of Solution

Justification:

It is known that a batch is characterized as acceptable if the proportion of defective components is at most 0.10 that is $p\le 0.10$.

Hence, the desirable probability of accepting a batch should be high when $p<0.10$ and the desirable probability of accepting a batch should be low when $p>0.10$

Comparing the three sampling plan, it can be observed that for p = 0.25, the probability of accepting a batch is lowest for sampling plan in part (d) and for p = 0.01, the probability of accepting a batch is highest for sampling plan in part (d).

Hence, the sampling plan in part (d) is most satisfactory.