Question

A manufacturer produces lots of a canned food product. Let p denote the proportion of the lots that do not meet the product quality specifications. An n = 33, c = 0 acceptance sampling plan will be used.

a) Compute points on the operating characteristic curve when

p = 0.01, 0.03, 0.10, and 0.20.

(Round your answers to four decimal places.)

Answer 1

Answer:-

Given that:-

The probability function for acceptance sampling is,

$f(x)=\frac{n!}{x!(n-x)!}p^{x}(1-p)^{(n-x)}$

where,

n= sample size

p=proportion of defective items in the lot

x=number of defective items in the sample

f(x)=Probability of x defective items in the smple.

(a)The probability of accepting a lot with n=33 and c=0 for 1%(p=0.01)defective is,

f(0) 33! 0!(33 – 0)! x (0.01)" x (1 – 0.01)(33–0).

  

33! 0!(33)! х (0.01) х (0.99)(33)

  

(0.99)

$=0.7177$

The probability of accepting a lot with n=33 and c=0 for 3%(p=0.03)defective is,

$f(0)=\frac{33!}{0!(33-0)!}\times (0.03)^{0}\times (1-0.03)^{(33-0)}$

  $=\frac{33!}{0!(33)!}\times (0.03)^{0}\times (0.97)^{(33)}$

  $=(0.97)^{33}$

  $=0.3659$

The probability of accepting a lot with n=33 and c=0 for 10%(p=0.10)defective is,

$f(0)=\frac{33!}{0!(33-0)!}\times (0.10)^{0}\times (1-0.10)^{(33-0)}$

$=\frac{33!}{0!(33)!}\times (0.10)^{0}\times (0.90)^{(33)}$

  $=(0.90)^{33}$

$=0.0309$

The probability of accepting a lot with n=33 and c=0 for 20%(p=0.20)defective is,

$f(0)=\frac{33!}{0!(33-0)!}\times (0.20)^{0}\times (1-0.20)^{(33-0)}$

$=\frac{33!}{0!(33)!}\times (0.20)^{0}\times (0.80)^{(33)}$

  $=(0.80)^{33}$

$=0.0006$

The probability of accepting a lot with n=33 and c=0 for different percentage of defectives are tabulated as:

percentage defective in the lot	probability of accepting the lot
1	0.7177
3	0.3659
10	0.0309
20	0.0006