The amount of time a bank teller spends with each customer has a population mean  = 3.10 minutes and standard deviation  = 0.40 minute. if a random sample of 16 customers is selected without replacement from a population of 500 customers, a. what is the probability that the average time spent per customer will be at least 3 minutes? b. there is an 85% chance that the sample mean will be below how many minutes?

a. First we solve the finite population correction factor σx from the formula:

σx = [s / sqrt(n)] * sqrt [(N – n) / (N – 1)]

σx = [0.40 / sqrt(16)] * sqrt[(500 – 16) / (500 – 1)]

σx = 0.0985

Then we compute for z score when x ≥ 3 minutes:

z = (x – m) / σx

z = (3 – 3.10) / 0.0985

z = -1.015

From the tables, the probability (p value) using right tailed test is:

P = 0.845 = 84.5%

b. At P = 0.85, the z score is z = 1.04

1.04 = (x – 3.10) / 0.0985

x = 3.20

Hence there is a 85% chance it will be below 3.20 minutes

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