Answer to Question #300964 in Statistics and Probability for jerie

Question

When purchasing bulk orders of​ batteries, a toy manufacturer uses this acceptance sampling​ plan: Randomly select and test 39 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 3000 ​batteries, and 1​% 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

let X represent the number of batteries

we define P( x≤3 ) = p(x=0 ) + p(x=1) + p(x=2) + p(x=3)

since we have been given that 0.01 do not meet specifications, then 0.99 meets specifications.

Then we find

p(x=0) = (0.99)³⁹ = 0.6757

p(x=1)= ( (0.01)¹(0.99)³⁸ ) * (39C1) = 0.2662

P(X=2)= ( (0.01)² (0.99)³⁷) ) * (39C2) = 0.05109

P(X=3) = ( (0.01)³ (0.99)³⁶ ) * (39C3) = 0.00636

Thus P( x≤3 ) = (0.6757 + 0.2662 + 0.05109 + 0.00636) = 0.99935 which is the required probability.

From the probability, we conclude that about 99.935% of all shipments will be accepted.