Question

PLZ HELP! Use technology or a z-score table to answer the question.

The amount spent by customers at a grocery store is normally distributed with a mean of $144.05 and a standard deviation of $34.52.

Approximately what percent of customers at the grocery store spend more than $200?


5.3%

8.2%

91.8%

94.7%

Answer 1

the answer would be A, or 5.3%

Explanation:

you start by finding the z-score of 200, so 200-144.05 = 55.95/34.52 = 1.62 (rounded to the nearest ) Then you use a z-score table to find the probability that a customer will spend less than 200 dollars and subtract the decimal probability from one. then you convert the decimal probability to a percent and get 5.3%.