When a truckload of apples arrives at a packing plant, a random sample of 150 is selected and examined for bruises, discoloration, and other defects. The whole truckload will be rejected if more than 5% of the sample is unsatisfactory. Suppose that in fact 8% of the apples on the truck do not meet the desired standard. What's the probability that the shipment will be accepted anyway?a. 0.0222b. 0.9778c. 0.088d. 0.912

Answer:c. 0.088Step-by-step explanation:Let p(s) be the proportion of apples defected in the sample. The probability that p(s)<0.05 can be calculated by calculating z statistic of 0.05:[tex]\frac{p(s)-p}{\sqrt{\frac{p*(1-p)}{N} } }[/tex] where p(s) = 0.05p is the proportion of the apples in fact defected (0.08)N is the sample size (150)   Then,z(0.05)=[tex]\frac{0.05-0.08}{\sqrt{\frac{0.08*0.92)}{150} } }=-1.354[/tex] And P(z<-1.354)≈0.0879Therefore the probability that  a random sample of 150 among 8% defected apples can be accepted is 0.088.