The amount of time a bank teller spends with each customer has a population mean

Question

The amount of time a bank teller spends with each customer has a population mean, mu, of 2.90 minutes and a standard deviation, sigma, of 0.40 minute. Complete parts (a) through (d). The amount of time If you select a random sample of 25 customers, what is the probability that the mean time spent per customer is at least 2.7 minutes? (Round to four decimal places as needed.) If you select a random sample of 25 customers, there is an 84% chance that the sample mean is less than how many minutes? (Round to four decimal places as needed.) What assumption must you make in order to solve (a) and (b)? That the population is symmetrically distributed such that the Central Limit Theorem w likely hold for samples of size 25 That the sample is symmetrically distributed such that the Central Limit Theorem w likely hold That the population is normally distributed That the sample is normally distributed That the population is uniformly distributed If you select a random sample of 64 customers, there is an 84% chance that the sample mean is less than how many minutes? (Round to four decimal places as needed.)

Explanation / Answer

1) std error =std deviation/(n)1/.2 =0.08

P(X>2.7)=1-P(Z<(2.7-2.9)/0.08)=1-P(Z<-2.5)=1-0.0062=0.9938

b) for 84 percentile; z=0.9945

hence correspondinfg value =2.9+0.9945*0.08=2.9796

c)option B

d) 2.9497

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