The average waiting time for a drive-in window at a local bank is 9.2 minutes, with a standard deviation of 2.6 minutes. When a customer arrives at the bank, find the probability that the customer will have to wait the given time. Assume the variable is normally distributed.(a) Between 5 and 10 minutes(b) Less than 6 minutes or more than 9 minutes

a) 56.91% probability that the customer will have to wait between 5 and 10 minutes.

b) 65.49% probability that the client will have to wait less than 6 minutes of more than 9 minutes.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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.

In this problem, we have that:

(a) Between 5 and 10 minutes

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 5. So

X = 10

has a pvalue of 0.6217

X = 5

has a pvalue of 0.0526

0.6217 - 0.0526 = 0.5691

56.91% probability that the customer will have to wait between 5 and 10 minutes.

(b) Less than 6 minutes or more than 9 minutes

Less than 6

pvalue of Z when X = 6

has a pvalue of 0.1230

12.30% probability that the client will have to wait less than 6 minutes

More than 9

1 subtracted by the pvalue of Z when X = 9.

has a pvalue of 0.4681

1 - 0.4681 = 0.5319

53.19% probability that the client will have to wait more than 9 minutes

Less than 6 or more than 9

12.30 + 53.19 = 65.49% probability that the client will have to wait less than 6 minutes of more than 9 minutes.

55? i think maybe

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