Answer:
A shopper that is at the 95th percentile of this distribution spends $116.125.
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 of $75 per trip and a standard deviation of $25 per trip.
This means that [tex]\mu = 75, \sigma = 25[/tex]
How much money does a shopper spend that is at the 95th percentile of this distribution?
This is X when Z has a p-value of 0.95, so X when Z = 1.645.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.645 = \frac{X - 75}{25}[/tex]
[tex]X - 75 = 1.645*25[/tex]
[tex]X = 116.125[/tex]
A shopper that is at the 95th percentile of this distribution spends $116.125.
