Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 53 batteries and determine whether each is within specifications. The entire shipment is accepted if at most 3 batteries do not meet specifications. A shipment contains 4000 ​batteries, and 2​% of them do not meet specifications. What is the probability that this whole shipment will be​ accepted? Will almost all such shipments be​ accepted, or will many be​ rejected?

Answer 1

Answer:

The probability of accepting this shipment is 98%.

Almost all such shipments will be accepted, because the probability of having a sample with more than 3 batteries defective is very low.

Step-by-step explanation:

The shipment will be accepted if at most 3 batteries don't meet specifications.

This can be modeled as a binomial distribution, where the sample size is n=53 an the probability of not meeting specifications of any battery is p=0.02.

The shipment will be accepted if $D\leq3$.

The probability of this event is:

$P(D\leq3)=P(D=0)+P(D=1)+P(D=2)+P(D=3)$

$P(D=k)=\frac{n!}{k!(n-k)!} p^k(1-p)^{n-k}\\\\\\P(D=0)=\frac{53!}{0!(53)!} 0.02^0*0.98^{53}=1*1*0.34=0.34\\\\P(D=1)=\frac{53!}{1!(52)!} 0.02^1*0.98^{52}=53*0.02*0.35=0.37\\\\P(D=2)=\frac{53!}{2!(51)!} 0.02^2*0.98^{51}=1378*0.0004*0.36=0.20\\\\P(D=3)=\frac{53!}{3!(50)!} 0.02^3*0.98^{50}=23426*0.000008*0.36=0.07$

Then,

$P(D\leq3)=P(D=0)+P(D=1)+P(D=2)+P(D=3)\\\\P(D\leq3)=0.34+0.37+0.20+0.07=0.98$

Answer 2

Answer:About 91.01 % will be accepted

Step-by-step explanation:

Given

probability of defective batteries is 2%

and sample tested is 53 batteries

p=0.02

n=53

Using binomial distribution

$P\left ( x\leq 2\right )=P\left ( x=0\right )+P\left ( x=1\right )+P\left ( x=2\right )$

$P\left ( x=0\right )=^{53}C_0\left ( 0.02\right )^0\left ( 0.98\right )&^53=0.3427$

$P\left ( x=1\right )=^{53}C_1\left ( 0.02\right )^1\left ( 0.98\right )^52=0.3707$

$P\left ( x=2\right )=^{53}C_2\left ( 0.02\right )^2\left ( 0.98\right )^51=0.19671$

$P\left ( x\leq 2\right )=0.3427+.3707+0.19671=0.91011$

$\approx 91.01 \%$