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 3000 ​batteries, and 3​% 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 is $P(X \le 2 ) = 0.9072$

The company will accept 90.72% of the shipment and will reject $(100 -90.72) = 9.2\%$ of the shipment , so many of the shipment are rejected

Step-by-step explanation:

From the question we are told that

   The  sample size is  n =  36

   The  proportion that did not meet the requirement is  $p = 0.03$

Generally the probability that the whole shipment is accepted is equivalent to the probability that there is at most 2 batteries that do not meet the requirement , this is mathematically represented as

     $P(X \le 2 ) = [ P(X = 0 ) + P(X = 1 ) + P(X = 0)]$

=>   $P(X \le 2 ) = [ [^{n}C_0 * (p)^{0} *(1-p)^{n-0} ] + [^{n}C_1 * (p)^{1} *(1-p)^{n-1} ] + [^{n}C_2 * (p)^{2} *(1-p)^{n-2} ]]$

Here C stands for Combination (so we will be making the combination function in our  calculators )

So

=> $P(X \le 2 ) = [ [^{36}C_0 * (0.03)^{0} *(1-0.03)^{36-0} ] + [^{36}C_1 * (0.03)^{1} *(1-0.03)^{36-1} ] + [^{36}C_2 * (0.03)^{2} *(1-0.03)^{36-2} ]]$

=> $P(X \le 2 ) = [ [1 * 1 * 0.3340 ] + [36* 0.03 *0.3444 ] + [630 * 0.0009 *(0.355 ]]$

=>$P(X \le 2 ) = 0.9072$

The company will accept 90.72% of the shipment and will reject $(100 -90.72) = 9.2\%$ of the shipment , so many of the shipment are rejected