store has deaided to accept a large shipment of ball-point pens if an inspection
ID: 3330007 • Letter: S
Question
store has deaided to accept a large shipment of ball-point pens if an inspection of 20 randomely selected pens yields no more (andthe probabilty that this shipment is accepted if 10% of the total shipment is defective. (use 3 decimal places.) (b) Find the probabilty that this ship ne tisnot accepted if 20% of the total shiprment is defective. (use 3 deci nal places.) than two dtedive . Can a normal distribution always be used to approximate a binomial distribution ? Explain your answer be accurate, np> 5 and ng > 5 O No. for the appreximation to O ves. We can ahnays use the normal distribution to approxsimate the binomial astribution. O No. For the approxdmation to be accurate, np>5 and ng 10 and ng > 10 Given that xisanormal variable with mean -43 and standard deviation ·7.1, nd the following probabilities. (Round your answers to four decimal places.) (a) Ptx s 60) (c) 50% x 60) 15-/2 pointsbBBeskcStat? 7 8.013 Find z such that 6% of the area under the standard normal curve les to the rgetofz. (Round your answer to two decimal places.) 1L One environmental group did a study of recyding habits in a California community. It found that 74% of the aluminum ans sold in the area were recycled. (Use the normal approximation. Round your answers to four decimal places.) n(a) If 395 cans are sold today, what is the probability that 300 or more will be recycled? -i.nl ""min.) (e (b) Of the 395 cans sold, what is the probability that between 260 and 300 will be recycled? Assume that 1Q scores are normally distributed, with a standard deviation of 20 points and a mean of 100 points. If 55 people are chosen at random, what is the probability that the sample mean of 1Q scores will not differ from the population mean by more than 2 points? (Round your answer to four decimal places.)Explanation / Answer
14.
We have been given params of the normal distribution :
Mean = 43
Stdev = 7.1
a. P(X<=60) = P(Z<= 60-43/ 7.1) = P(Z<=2.4) = .9918
b. P(X>=60) = P(Z>= 60-43/ 7.1) = P(Z>=2.4) = 1-.9918 = .0082
c. P(50<=X<=60) = P(.99<=Z<=2.4) = .9918-.8389 = .1529
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