Chapter 12, Problem 67AP

a

Summary Introduction

Interpretation:

Acceptability of heavy duty nails lot

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Answer to Problem 67AP

  $\alpha$ is 0.0803 and $\beta$ is 0.0028.

Explanation of Solution

Given information:

n = 100

c= 2

AQL = 1%

LTPD = 10%

  $\alpha =0.0803$ as = 10000

and $\beta =0.0028$ are the producers and consumers risk factor respectively. $\alpha$ , represents the probability that a sampling plan might reject the good lot whereas $\beta$ , represents the probability that a sampling plan might accept the bad lot.

Lot size, N=10,000

Accepted Quality Level (AQL) is 1%

Lot Tolerance Percent Defective (LTPD) is 10%

  $\begin{array}{l}{p}_o=0.01\\ p_1=0.10\end{array}$

  $\alpha =P\left\{X>c/p=1,n=100\right\}$

  $=P\left\{X>2/\theta =1\right\}=0.0803$

  $\beta =P\left\{X\pounds /p=0.1,n=100\right\}$

  $=P\left\{X\pounds 2/\theta =10\right\}=0.0028$

Hence, $\alpha$ is 0.0803 and $\beta$ is 0.0028.

b

Summary Introduction

Interpretation:

Sequential sampling plan

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Answer to Problem 67AP

Sequential sampling plan with two limit lines are $x_a=-2.42+0.0394n$ and $x_t=1.051+0.0397n$

Explanation of Solution

Given information:

n = 100

c= 2

AQL = 1%

LTPD = 10%

  $\alpha =0.0803$ as = 10000

Computation of two lines

Calculate the values of k, $h_1$ and $h_2$ by using the following formula.

  $k=\log \left[\frac{p_2\left(1-p_1\right)}{p_1\left(1-p_2\right)}\right]\mathrm{.....}\left(1\right)$

  $h_1=\frac{1}{k}\left[\log \left(\frac{1-\alpha }{\beta }\right)\right]\mathrm{.....}\left(2\right)$

  $h_2=\frac{1}{k}\left[\log \left(\frac{1-\beta }{\alpha }\right)\right]\mathrm{......}\left(3\right)$

  $s=\frac{1}{k}\left[\log \left(\frac{1-0.01}{1-0.10}\right)\right]\mathrm{....}\left(4\right)$

Information known as per the problem

  $\begin{array}{l}{p}_1=0.01\\ p_2=0.10\\ \alpha =0.0803\\ \beta =0.0028\end{array}$

Now, substitute these values in equation (1) to calculate k as shown below

  $k=\log \left[\frac{p_2\left(1-p_1\right)}{p_1\left(1-p_2\right)}\right]$

  $=\log \left[\frac{0.10\left(1-0.01\right)}{0.01\left(1-0.10\right)}\right]$

  $=\log \frac{0.099}{0.009}=\log 11=1.041$

Now, substitute these values in equation (2) to calculate $h_1$ as shown below:

  $h_1=\frac{1}{k}\left[\log \left(\frac{1-\alpha }{\beta }\right)\right]$

  $=\frac{1}{1.041}\left[\log \left(\frac{1-0.0803}{0.0028}\right)\right]$

  $=\frac{1}{1.041}\left[\log \left(\frac{0.9197}{0.0028}\right)\right]$

  $=0.96\left(\log 328.46\right)=0.96\times 2.52=2.42$

Now, substitute these values in equation (3) to calculate $h_2$ as shown below:

  $h_2=\frac{1}{k}\left[\log \left(\frac{1-\beta }{\alpha }\right)\right]$

= $\frac{1}{0.041}\left[\log \left(\frac{1-0.0028}{0.0803}\right)\right]$

  $\begin{array}{l}=\frac{1}{1.041}\left[\log \left(\frac{0.9972}{0.0803}\right)\right]\\ =0.96\left(\log 12.42\right)=0.96\times 1.095=1.051\\ \end{array}$

Similarly, substitute these values in equation (4) to calculate s as shown below:

  $s=\frac{1}{k}\left[\log \left(\frac{1-0.01}{1-0.10}\right)\right]$

  $=\frac{1}{1.041}\left[\log \left(\frac{0.99}{0.90}\right)\right]$

  $=0.96\left[\log \left(1.1\right)\right]$

  $=0.96\times 0.0413=0.0397$

Calculate the two limit lines by substituting the values of 2.42, 1.051 and 0.0397 in the $h_1$ , $h_2$ and s respectively in the below formulae.

  $x_a=-h_1+sn$

  $x_t=h_2+sn$

  $x_a=-2.42+0.0394n$

  $x_t=1.051+0.0397n$

Hence, the two limit lines are $x_a=-2.42+0.0394n$ and $x_t=1.051+0.0397n$ .

c

Summary Introduction

Interpretation:

Maximum value of the expected sample size

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Answer to Problem 67AP

The maximum expected number of items which must be sampled is probably less than 75.

Explanation of Solution

Given information:

n = 100

c= 2

AQL = 1%

LTPD = 10%

  $\alpha =0.0803$ as = 10000

At p=0 ASN=60.86

At p=0.1 ASN=71.97

At p=0.0397 ASN=66.6

Based on this, the maximum expected number of items which must be sampled is probably less than 75.

d

Summary Introduction

Interpretation:

Graphical representation of acceptance and rejection region and check whether sequential sampling plan would recommend acceptance or rejection on or before testing 100^th nail.

Concept Introduction:

Normal distribution is the probability function with continuous series. It is bell shaped distribution function where mean, median and mode are same.

Explanation of Solution

  

Hence, acceptance would be recommended on 61^st nail.