Chapter 19, Problem 15E

Determine the probability of accepting lots that are 10%, 20%, 30%, and 40% defective using a sample of size 12 and an acceptance number of 2.

To determine

Find the probability of accepting lots that are 10%, 20%, 30%, and 40% defective.

Answer to Problem 15E

The probability of accepting lots that are 10%, 20%, 30%, and 40% defective is,

Defective Percent	Probability of accepting lot
10	0.889
20	0.558
30	0.253
40	0.083

Explanation of Solution

Calculation:

Let x denotes the accepting lots.

For 10% defective:

The probability of accepting lots that is 10% defective is,

$P\left(x\le 2|\pi =0.10\text{ and }n=12\right)=\left\{\begin{array}{l}P\left(x=0|\pi =0.10\text{ and }n=12\right)+\\ P\left(x=1|\pi =0.10\text{ and }n=12\right)+\\ P\left(x=2|\pi =0.10\text{ and }n=12\right)\end{array}\right\}$

Compute the probability values for each x using binomial probability distribution table.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.10 in probability row.
• Locate the value 0 in x column.
• The intersecting value that corresponds to 0 with probability 0.10 is 0.282.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.10 in probability row.
• Locate the value 1 in x column.
• The intersecting value that corresponds to 1 with probability 0.10 is 0.377.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.10 in probability row.
• Locate the value 2 in x column.
• The intersecting value that corresponds to 2 with probability 0.10 is 0.230.

The required probability is,

$\begin{array}{c}P\left(x\le 2|\pi =0.10\text{ and }n=12\right)=0.282+0.377+0.230\\ =0.889\end{array}$

Hence, the probability of accepting lots that is 10% defective is 0.889.

For 20% defective:

The probability of accepting lots that is 20% defective is,

$P\left(x\le 2|\pi =0.20\text{ and }n=12\right)=\left\{\begin{array}{l}P\left(x=0|\pi =0.20\text{ and }n=12\right)+\\ P\left(x=1|\pi =0.20\text{ and }n=12\right)+\\ P\left(x=2|\pi =0.20\text{ and }n=12\right)\end{array}\right\}$

Compute the probability values for each x using binomial probability distribution table.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.20 in probability row.
• Locate the value 0 in x column.
• The intersecting value that corresponds to 0 with probability 0.20 is 0.069.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.20 in probability row.
• Locate the value 1 in x column.
• The intersecting value that corresponds to 1 with probability 0.20 is 0.206.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.20 in probability row.
• Locate the value 2 in x column.
• The intersecting value that corresponds to 2 with probability 0.20 is 0.283.

The required probability is,

$\begin{array}{c}P\left(x\le 2|\pi =0.20\text{ and }n=12\right)=0.069+0.206+0.283\\ =0.558\end{array}$

Hence, the probability of accepting lots that is 20% defective is 0.558.

For 30% defective:

The probability of accepting lots that is 30% defective is,

$P\left(x\le 2|\pi =0.30\text{ and }n=12\right)=\left\{\begin{array}{l}P\left(x=0|\pi =0.30\text{ and }n=12\right)+\\ P\left(x=1|\pi =0.30\text{ and }n=12\right)+\\ P\left(x=2|\pi =0.30\text{ and }n=12\right)\end{array}\right\}$

Compute the probability values for each x using binomial probability distribution table.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.30 in probability row.
• Locate the value 0 in x column.
• The intersecting value that corresponds to 0 with probability 0.30 is 0.014.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.30 in probability row.
• Locate the value 1 in x column.
• The intersecting value that corresponds to 1 with probability 0.30 is 0.071.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.30 in probability row.
• Locate the value 2 in x column.
• The intersecting value that corresponds to 2 with probability 0.30 is 0.168.

The required probability is,

$\begin{array}{c}P\left(x\le 2|\pi =0.20\text{ and }n=12\right)=0.014+0.071+0.168\\ =0.253\end{array}$

Hence, the probability of accepting lots that is 30% defective is 0.253.

For 40% defective:

The probability of accepting lots that is 40% defective is,

$P\left(x\le 2|\pi =0.40\text{ and }n=12\right)=\left\{\begin{array}{l}P\left(x=0|\pi =0.40\text{ and }n=12\right)+\\ P\left(x=1|\pi =0.40\text{ and }n=12\right)+\\ P\left(x=2|\pi =0.40\text{ and }n=12\right)\end{array}\right\}$

Compute the probability values for each x using binomial probability distribution table.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.40 in probability row.
• Locate the value 0 in x column.
• The intersecting value that corresponds to 0 with probability 0.40 is 0.002.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.40 in probability row.
• Locate the value 1 in x column.
• The intersecting value that corresponds to 1 with probability 0.40 is 0.017.

From the Appendix B: Tables B.1 Binomial Probability Distribution:

• Select the table with sample size 12.
• Locate the value 0.40 in probability row.
• Locate the value 2 in x column.
• The intersecting value that corresponds to 2 with probability 0.40 is 0.064.

The required probability is,

$\begin{array}{c}P\left(x\le 2|\pi =0.20\text{ and }n=12\right)=0.002+0.017+0.064\\ =0.083\end{array}$

Hence, the probability of accepting lots that is 40% defective is 0.083.