Question

Question Help When purchasing bulk orders of batteries, a toy manufacturer uses this acceptance sampling plan: Randomly select and test 36 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 2 batteries do not meet specifications. A shipment contains 4000 batteries, and 1% of them do not meet specifications. What is the probability that this whole shipment will be accepted? Wil almost all such shipments be accepted, or will many be rejected? The probability that this whole shipment will be accepted is (Round to four decimal places as needed.) % of the shipments, so The company will accept % of the shipments and will reject (Round to two decimal places as needed.)

Answer 1

Answer:

Step-by-step explanation:

Here X no of defective batteries is binomial with p = 1% = 0.01

q = 1-p=0.99

(Because each battery is independent and there are only two outcomes)

Prob that shipment is accepted is for n =36, $P(X\leq 2)$

=P(x=0,1,2)

= $\Sigma^2_0 36Cr (0.01)^r (0.99)^{36-r}$

= 0.9944

Answer 2

Answer:

The probability of accepting this shipment is 0.9944.

The company will accept 99.44% of the shipments and reject 0.66% of the shipments.

Step-by-step explanation:

The shipment will be accepted if at most 2 batteries do noy meet the specifications in the sample of 36 batteries.

We can model this as a binomial distribution, with sample size of n=36 and probability of defective of p=0.01.

The probability of having at most 2 batteries defective is:

$P(d\leq2)=P(d=0)+P(d=1)+P(d=2)\\\\\\P(d=k)=\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}\\\\\\ P(d=0)=\frac{36!}{0!36!}*0.01^0*0.99^{36}=1*1+0.6964=0.6964\\\\P(d=1)=\frac{36!}{1!35!}*0.01^1*0.99^{35}=36*0.01+0.7034=0.2532\\\\P(d=2)=\frac{36!}{2!34!}*0.01^2*0.99^{34}=630*0.0001+0.7106=0.0448\\\\\\P(d\leq2)=P(d=0)+P(d=1)+P(d=2)\\\\P(d\leq2)=0.6964+0.2532+0.0448=0.9944$