Question

A company purchases shipments of machine components and uses this acceptance sampling​ plan: Randomly select and test 2626 components and accept the whole batch if there are fewer than 3 defectives. If a particular shipment of thousands of components actually has a 77​% rate of​ defects, what is the probability that this whole shipment will be​ accepted? Round to three decimal places.

Answer 1

Question is not well presented;

Question;

A company purchases shipments of machine components and uses this acceptance sampling​ plan: Randomly select and test 26 components and accept the whole batch if there are fewer than 3 defectives. If a particular shipment of thousands of components actually has a 7% rate of​ defects, what is the probability that this whole shipment will be​ accepted? Round to three decimal places.

Answer:

0.727

Step-by-step explanation:

Given

n = 26

Let P(D) = Probability of defection

P(D) = 7% = 0.07

The whole batch are to be accepted, if there are fewer than 3 defectives; so, we are looking for P(0≤x≤2).

This is binomial.

Solving the above binomial problem;

We have

P(0≤x≤2) = binomcdf(26,0.07,2)

P(0≤x≤2) = 0.7272

P(0≤x≤2) = 0.727 --- Approximated

Answer 2

Answer:

Probability that whole shipment will be​ accepted is $9.48 \times 10^{-14}$ .

Step-by-step explanation:

We are given that a company purchases shipments of machine components and uses this acceptance sampling​ plan: Randomly select and test 26 components and accept the whole batch if there are fewer than 3 defectives. It is known that a particular shipment of thousands of components actually has a 77​% rate of​ defects.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 26 components

            r = number of success = fewer than 3 defects

           p = probability of success which in our question is % rate of defects,

                   i.e., 77%

LET X = Number of defective items

Also, it is given that a sample of 26 items is taken,

So, it means X ~ $Binom(n=26,p=0.77)$

(a) Probability that the inspection procedure will pass the shipment is given by the fact that if fewer than 3 components found to be defective, then only shipment is passed, i.e.;

P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{26}{0}0.77^{0} (1-0.77)^{26-0} + \binom{26}{1}0.77^{1} (1-0.77)^{26-1}+\binom{26}{2}0.77^{2} (1-0.77)^{26-2}$            

= $1 \times 1 \times 0.23^{26} +26 \times 0.77 \times 0.23^{25}+325 \times 0.77^{2} \times 0.23^{24}$

= $9.48 \times 10^{-14}$

Therefore, probability that whole shipment will be​ accepted is $9.48 \times 10^{-14}$ .