Question

A company purchases shipments of machine components and uses this acceptance 11) sampling plan: Randomly select and test 30 components and accept the whole batch if there are fewer than 3 defectives. If a particular shipment of thousands of components actually has a 6 % rate of defects, what is the probability that this whole shipment will be accepted?

Answer 1

Answer:

Probability that this whole shipment will be accepted is 0.7324.

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 30 components and accept the whole batch if there are fewer than 3 defectives.

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 = 30 components

            r = number of success = fewer than 3

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

                 of defects, i.e; 6%

LET X = Number of defective components

So, it means X ~ $Binom(n=30, p=0.06)$

Now, Probability that this whole shipment will be accepted only when there are fewer than 3 defectives = P(X < 3)

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

= $\binom{30}{0}\times 0.06^{0} \times (1-0.06)^{30-0}+ \binom{30}{1}\times 0.06^{1} \times (1-0.06)^{30-1}+ \binom{30}{2}\times 0.06^{2} \times (1-0.06)^{30-2}$

= $1 \times 1 \times 0.94^{30}+ 30 \times 0.06^{1} \times 0.94^{29}+ 435 \times 0.06^{2} \times 0.94^{28}$

= 0.7324

Therefore, probability that this whole shipment will be accepted is 0.7324.