Answer:
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.
