Question

A manufacturer receives a lot of 270 parts from a vendor. The lot will be unacceptable if more than 12 of the parts are defective. The manufacturer is going to select randomly k parts from the lot for inspection and the lot will be accepted if no defective parts are found in the sample. • (a) How large does K have to be to ensure that the probability that the manufacturer accepts an unacceptable lot is less than 0.20? • (b) How large does K have to be to ensure that the probability that there are exactly two defectives in the sample is less than 0.20? . (c) Suppose the manufacturer decides to accept the lot if there are at most two defectives in the sample. How large does K have to ensure that the probability that manufacturer accepts an unacceptable lot is less than 0.20?

Answer 1

Answer:-

Given That:-

A manufacturer receives a lot of 270 parts from a vendor. The lot will be unacceptable if more than 12 of the parts are defective. The manufacturer is going to select randomly k parts from the lot for inspection and the lot will be accepted if no defective parts are found in the sample. •

A manufacture receive a lot of 270 parts from a vendor The lot will be unaceptable if more than 12 of the parts are defctive.

The manufacture going to select k smaple from 270 parts.

Let X be random variable reopresent number of defective items in the sample.

Therefore,

X ~ B(n, p)

Where

n = k = sample size

p = probability of defective items in the lot.

(a) How large does K have to be to ensure that the probability that the manufacturer accepts an unacceptable lot is less than 0.20? •

We want to find k = ?

When probability that the manufacture accepts an unacceptable lot is less than 0.20

Let is unaccepted if defectives > 12

Probability of defective = 13/270

= 0.0481 to 1

0.0481 < p < 1

Minimum probability of defective p = 0.0481

Since manufacture selected k sample size

Then

p[No defective] < 0.20

p[x = 0] < 0.20

Because X ~ B(k, p)

K 1-YI 9 <0.20

$\binom{k}{0} q^{k}<0.20$

< 0.2

$(1-0.0481)^k<0.2$

Use R command

k : 0 to 270

p = $(1-0.0481)^k$ : 1 to 0.00001

for k > 34 $(1-0.0481)^k$ < 0.2

k = 34 is the minimum sample size for which given condition satisfied.

sample size k = 34

(b) How large does K have to be to ensure that the probability that there are exactly two defectives in the sample is less than 0.20?

k = ?

For probability that there are exactly two defectives in the sample is less than 0.20

i.e,

p[X = 2] < 0.20

$\binom{k}{2}(0.0481)^2(1-0.0481)^{k-2}<0.20$

R common

k : 0 to 270

$\binom{k}{2}(0.0481)^2(1-0.0481)^{k-2}=1\, \, to$

k < 172.544238

c) Suppose the manufacturer decides to accept the lot if there are at most two defectives in the sample. How large does K have to ensure that the probability that manufacturer accepts an unacceptable lot is less than 0.20?

k = ?

Defcetives at most two

$p[X\leq 2]<0.2$

$\sum_{x=0}^{2}\binom{k}{x}p^xq^{k-x}<0.2$

Therefore k > 89

89 < k < 270 p < 0.2

Minimum sample size is k = 89 for probability < 0.2

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