Question

The amount of time that a drive-through bank teller spends on a customer is a random variable with a mean u = 7.9 minutes and a standard deviation o = 3.6 minutes. If a random sample of 81 customers is observed, find the probability that their mean time at the teller's window is (a) at most 7.3 minutes; (b) more than 8.7 minutes; (c) at least 7.9 minutes but less than 8.3 minutes. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. (a) The probability that the mean time is at most 7.3 minutes is (Round to four decimal places as needed.) (b) The probability that the mean time is more than 8.7 minutes is (Round to four decimal places as needed.) (C) The probability that the mean time is between 7.9 minuts and 8.3 minutes is (Round to four decimal places as needed.) . Enter your answer in each of the answer boxes.

Answer 1

Solution. - Given that, u = 0 = n 7.9 3.6 81 Now, My al = 7.9 od .04 a. Probability PCX (73) P(x < 7.3) =P / X-My 7.3 - 7.9 0.4 - P(Z < -1.50) = 0.0668 - Using 2-table. P - | Probability P(x 78.7) P(X 78.7) = 1- P(X L8.7) - 1-P / X-Mx 1 or 8.9-79) 0.4 : 1 - P(Z < 2.00) 1 - 0.9772 — = 0.0228 Using 2-table

c Probability PL 7.9 < X < 8.3) - P(X < 8.3) - P(X < 7.9) = P/x-My 8.3-7.9 p/ X-lx 7.9-7.9 I or 0.4 / or 0.4 P(Z < 1.00) - P(Z < 0.00) : 0.8413 - 0.5000 - Using z-table ooo 0.3413