A manufacturer of computer printers purchases plastic ink cartridges from a vendor. When a large shipment is received, a random sample of 230 cartridges is selected, and each cartridge is inspected. If the sample proportion of defective cartridges is more than 0.02, the entire shipment is returned to the vendor. (a) What is the approximate probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is 0.06? (Round your answer to four decimal places.) (b) What is the approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.10? (Round your answer to four decimal places.) 0

Answer:a) There is a 99.48% probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is 0.06.b) There is a 0% probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.10.Step-by-step explanation:For eah cartridge, there are only two possible outcomes. Either it is defective, or it is not. This means that we are going to use the binomial probability distribution to solve this problem.However, we are working with samples that are considerably big. So i am going to aaproximate this binomial distribution to the normal.Binomial probability distributionProbability of exactly x sucesses on n repeated trials, with p probability.Can be approximated to a normal distribution, using the expected value and the standard deviation.The expected value of the binomial distribution is:[tex]E(X) = np[/tex]The standard deviation of the binomial distribution is:[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]Normal probability distributionProblems of normally distributed samples can be solved using the z-score formula.In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:[tex]Z = \frac{X - \mu}{\sigma}[/tex]The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.When we are approximating a binomial distribution to a normal one, we have that [tex]\mu = E(X), \sigma = \sqrt{V(X)}[/tex]In this problem, we have that:There are 230 cartridges, so [tex]n = 230[/tex].If there are more than 0.02*230 = 4.6 defective, the sample will be returned.(a) What is the approximate probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is 0.06?In this case, we have that:[tex]\mu = E(X) = 0.06*230 = 13.8[/tex][tex]\sigma = \sqrt{V(X)} = \sqrt{230*0.06*0.94} = 3.60[/tex]This probability is 1 subtracted by the pvalue of Z when [tex]X = 4.6[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{4.6 - 13.8}{3.60}[/tex][tex]Z = -2.56[/tex][tex]Z = -2.56[/tex] has a pvalue of 0.0052This means that there is a 1-0.0052 = 0.9948 = 99.48% probability that a shipment will be returned if the true proportion of defective cartridges in the shipment is 0.06.(b) What is the approximate probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.10? In this case, we have that:[tex]\mu = E(X) = 0.10*230 = 23[/tex][tex]\sigma = \sqrt{V(X)} = \sqrt{230*0.10*0.90} = 4.55[/tex]This probability is the pvalue of Z when [tex]X = 4.6[/tex][tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]Z = \frac{4.6 - 23}{4.6}[/tex][tex]Z = -4[/tex][tex]Z = -4[/tex] has a pvalue of 0.This means that there is a 0% probability that a shipment will not be returned if the true proportion of defective cartridges in the shipment is 0.10.