Question

A quality control engineer is in charge of testing whether or not 90% of the DVD players produced by his company conform to specifications. To do this, the engineer randomly selects a batch of 12 DVD players from each day's production. The day's production is acceptable provided no more than 1 DVD player fails to meet specifications. Otherwise, the entire day's production has to be tested. (i) What is the probability that the engineer incorrectly passes a day's production as acceptable if only 80% of the day's DVD players actually conform to specification? (2 points) (ii) What is the probability that the engineer unnecessarily requires the entire day's production to be tested if in fact 90% of the DVD players conform to specifications? (2 points)

Answer 1

The solution is done showing all steps with necessary calculations and required formulas. Hope you can understand.

Let y represents the number of DVD players in the sample that fail to meet specifications. we will use Binomial distribution approach, which is given by : n = total trials. prm) p* (1-0) ; x no. of success pa probability of success M :) Here, n=12, p=1-80 l 80% of the day's DVD players actually 100 L conform to specifications | 0.2 we have to find PCXE1) PCX2 ) - P2 - ) + PC > = ) = (3) 60.29C1-0.2)2+(-3) (0-2)' (1-0.27" = 0.2748 (Aus) 5) Here, p = 1- 9o =0.1 n = 12 We have to find P(X>1) here. PCs > ) = (- P(x< 1) [ f(> >a) = |- P(x áo). - 1- [PC ) + (x =)) 31-[(3)(0.1°1-0.1)*7 (13)60-17"CI-0.0"] = -0,659 = 0.341 (Aus)