According to loyalty card data, the amount of money spent per shopper at the grocery store is normally distributed with a mean of $75 per trip and a standard deviation of $25 per trip. How much money does a shopper spend that is at the 95th percentile of this distribution

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 and standard deviation , the z-score 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 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

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.

A shopper that is at the 95th percentile of this distribution spends $116.125.

answer: the answer is 127 degrees

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)