Chapter 3, Problem 183SE

a

To determine

Identify the sampling plan that can be preferred, if the seller produces lots with fraction defective ranging from $p=0$ and $p=0.10$.

Answer to Problem 183SE

The sampling plan that can be preferred, if the seller produces lots with fraction defective ranging from $p=0$ and $p=0.10$ is sampling plan with $n=25,a=5$.

Explanation of Solution

Calculation:

Binomial distribution:

A random variable Y is a binomial distribution based on n trails with success probability p if and only if,

$p\left(y\right)=\left(\begin{array}{l}n\\ y\end{array}\right)p^y{q}^{n-y},y=0,1,2,\cdots ,n\text{ and }0\le p\le 1$

For sampling plan 1 with $n=5,a=1$:

The probability of acceptance for $p=0.05$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left(\begin{array}{l}5\\ 0\end{array}\right){\left(0.05\right)}^0{\left(1-0.05\right)}^{5-0}+\left(\begin{array}{l}5\\ 1\end{array}\right){\left(0.05\right)}^1{\left(1-0.05\right)}^{5-1}\\ =0.7738+0.2036\\ =0.9774\end{array}$

The probability of acceptance for $p=0.10$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left(\begin{array}{l}5\\ 0\end{array}\right){\left(0.10\right)}^0{\left(1-0.10\right)}^{5-0}+\left(\begin{array}{l}5\\ 1\end{array}\right){\left(0.10\right)}^1{\left(1-0.10\right)}^{5-1}\\ =0.5905+0.3281\\ =0.9186\end{array}$

The probability of acceptance for $p=0.20$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left(\begin{array}{l}5\\ 0\end{array}\right){\left(0.20\right)}^0{\left(1-0.20\right)}^{5-0}+\left(\begin{array}{l}5\\ 1\end{array}\right){\left(0.20\right)}^1{\left(1-0.20\right)}^{5-1}\\ =0.3277+0.4096\\ =0.7373\end{array}$

The probability of acceptance for $p=0.30$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left(\begin{array}{l}5\\ 0\end{array}\right){\left(0.30\right)}^0{\left(1-0.30\right)}^{5-0}+\left(\begin{array}{l}5\\ 1\end{array}\right){\left(0.30\right)}^1{\left(1-0.30\right)}^{5-1}\\ =0.1681+0.3602\\ =0.5283\end{array}$

The probability of acceptance for $p=0.40$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left(\begin{array}{l}5\\ 0\end{array}\right){\left(0.40\right)}^0{\left(1-0.40\right)}^{5-0}+\left(\begin{array}{l}5\\ 1\end{array}\right){\left(0.40\right)}^1{\left(1-0.40\right)}^{5-1}\\ =0.0778+0.2592\\ =0.3370\end{array}$

The probability of acceptance for sampling plan $n=5,a=1$ is,

p	0.05	0.10	0.20	0.30	0.40
$p\left(\text{acceptance}\right)$	0.9774	0.9186	0.7373	0.5283	0.3370

For sampling plan 2 with $n=25,a=5$:

The probability of acceptance for $p=0.05$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left[\begin{array}{c}\left(\begin{array}{l}25\\ 0\end{array}\right){\left(0.05\right)}^0{\left(1-0.05\right)}^{25-0}+\left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.05\right)}^1{\left(1-0.05\right)}^{25-1}+\left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.05\right)}^2{\left(1-0.05\right)}^{25-2}+\\ \left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.05\right)}^3{\left(1-0.05\right)}^{25-3}+\left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.05\right)}^4{\left(1-0.05\right)}^{25-4}+\left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.05\right)}^5{\left(1-0.05\right)}^{25-5}\end{array}\right]\\ =0.2774+0.3650+0.2305+0.0930+0.0269+0.0059\\ =0.9987\end{array}$

The probability of acceptance for $p=0.10$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left[\begin{array}{c}\left(\begin{array}{l}25\\ 0\end{array}\right){\left(0.10\right)}^0{\left(1-0.10\right)}^{25-0}+\left(\begin{array}{l}25\\ 1\end{array}\right){\left(0.10\right)}^1{\left(1-0.10\right)}^{25-1}+\left(\begin{array}{l}25\\ 2\end{array}\right){\left(0.10\right)}^2{\left(1-0.10\right)}^{25-2}+\\ \left(\begin{array}{l}25\\ 3\end{array}\right){\left(0.10\right)}^3{\left(1-0.10\right)}^{25-3}+\left(\begin{array}{l}25\\ 4\end{array}\right){\left(0.10\right)}^4{\left(1-0.10\right)}^{25-4}+\left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.10\right)}^5{\left(1-0.10\right)}^{25-5}\end{array}\right]\\ =0.0718+0.1994+0.2659+0.2265+0.1384+0.0646\\ =0.9666\end{array}$

The probability of acceptance for $p=0.20$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left[\begin{array}{c}\left(\begin{array}{l}25\\ 0\end{array}\right){\left(0.20\right)}^0{\left(1-0.20\right)}^{25-0}+\left(\begin{array}{l}25\\ 1\end{array}\right){\left(0.20\right)}^1{\left(1-0.20\right)}^{25-1}+\left(\begin{array}{l}25\\ 2\end{array}\right){\left(0.20\right)}^2{\left(1-0.20\right)}^{25-2}+\\ \left(\begin{array}{l}25\\ 3\end{array}\right){\left(0.20\right)}^3{\left(1-0.20\right)}^{25-3}+\left(\begin{array}{l}25\\ 4\end{array}\right){\left(0.20\right)}^4{\left(1-0.20\right)}^{25-4}+\left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.20\right)}^5{\left(1-0.20\right)}^{25-5}\end{array}\right]\\ =0.0038+0.0236+0.0708+0.1356+0.1867+0.1960\\ =0.6165\end{array}$

The probability of acceptance for $p=0.30$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left[\begin{array}{c}\left(\begin{array}{l}25\\ 0\end{array}\right){\left(0.30\right)}^0{\left(1-0.30\right)}^{25-0}+\left(\begin{array}{l}25\\ 1\end{array}\right){\left(0.30\right)}^1{\left(1-0.30\right)}^{25-1}+\left(\begin{array}{l}25\\ 2\end{array}\right){\left(0.30\right)}^2{\left(1-0.30\right)}^{25-2}+\\ \left(\begin{array}{l}25\\ 3\end{array}\right){\left(0.30\right)}^3{\left(1-0.30\right)}^{25-3}+\left(\begin{array}{l}25\\ 4\end{array}\right){\left(0.30\right)}^4{\left(1-0.30\right)}^{25-4}+\left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.30\right)}^5{\left(1-0.30\right)}^{25-5}\end{array}\right]\\ =0.0001+0.0014+0.0074+0.0243+0.0572+0.1030\\ =0.1934\end{array}$

The probability of acceptance for $p=0.40$ is,

$\begin{array}{c}p\left(\text{acceptance}\right)=\left[\begin{array}{c}\left(\begin{array}{l}25\\ 0\end{array}\right){\left(0.40\right)}^0{\left(1-0.40\right)}^{25-0}+\left(\begin{array}{l}25\\ 1\end{array}\right){\left(0.40\right)}^1{\left(1-0.40\right)}^{25-1}+\left(\begin{array}{l}25\\ 2\end{array}\right){\left(0.40\right)}^2{\left(1-0.40\right)}^{25-2}+\\ \left(\begin{array}{l}25\\ 3\end{array}\right){\left(0.40\right)}^3{\left(1-0.40\right)}^{25-3}+\left(\begin{array}{l}25\\ 4\end{array}\right){\left(0.40\right)}^4{\left(1-0.40\right)}^{25-4}+\left(\begin{array}{l}25\\ 5\end{array}\right){\left(0.40\right)}^5{\left(1-0.40\right)}^{25-5}\end{array}\right]\\ =0.000003+0.000047+0.00038+0.00194+0.00710+0.01989\\ =0.02936\end{array}$

The probability of acceptance for sampling plan $n=25,a=5$ is,

p	0.05	0.10	0.20	0.30	0.40
$p\left(\text{acceptance}\right)$	0.9987	0.9666	0.6165	0.1934	0.02936

Step by step procedure to construct OC curve:

• In OC curve, take the values of probability p on x-axis.
• Take the values of probability of acceptance for p on y-axis.
• Locate the value of probability of acceptance 0.0059 corresponding to probability 0.05 for sampling plan 1.
• Similarly locate all the probability values for both the sampling plans.
• Connect all the dots with a curved line to form the OC curve.

The OC curve is,

From the OC curve it can be observed that, for the interval $p=0$ and $p=0.10$, the line for sampling plan 2 is above the line of sampling plan 1. This shows that, using sampling plan 2 would be preferable if the seller produces lots with fraction defective ranging from $p=0$ and $p=0.10$.

Hence, the sampling plan that can be preferred, if the seller produces lots with fraction defective ranging from $p=0$ and $p=0.10$ is sampling plan 2 with $n=25,a=5$.

b

To determine

Identify the sampling plan that can be preferred, if the seller produces lots with fraction defective exceeding $p=0.30$.

Answer to Problem 183SE

The sampling plan that can be preferred, if the seller produces lots with fraction defective exceeding $p=0.30$ is sampling plan 2 with $n=25,a=5$.

Explanation of Solution

From the OC curve of part (a), it can be observed that, for the p values exceeding 0.30, the line for sampling plan 2 is above the line of sampling plan 1. This shows that, using sampling plan 2 would be preferable if the seller produces lots with fraction defective exceeding 0.30.

Hence, the sampling plan that can be preferred, if the seller produces lots with fraction defective exceeding $p=0.30$ is sampling plan 2 with $n=25,a=5$.