Respuesta :
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;
[tex]P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....[/tex]
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 ~ [tex]Binom(n=30, p=0.06)[/tex]
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)
= [tex]\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}[/tex]
= [tex]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}[/tex]
= 0.7324
Therefore, probability that this whole shipment will be accepted is 0.7324.
