Answer:
5.48% of the people in line waited for more than 28 minutes
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 p-value, we get the probability that the value of the measure is greater than X.
Mean waiting time of 20 minutes with a standard deviation of 5 minutes.
This means that [tex]\mu = 20, \sigma = 5[/tex]
What percentage of the people in line waited for more than 28 minutes?
The proportion is 1 subtracted by the p-value of Z when X = 28. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{28 - 20}{5}[/tex]
[tex]Z = 1.6[/tex]
[tex]Z = 1.6[/tex] has a p-value of 0.9452.
1 - 0.9452 = 0.0548.
As a percentage:
0.0548*100% = 5.48%
5.48% of the people in line waited for more than 28 minutes
