Answer:
0.02% probability that the average amount billed on the sample bills is greater than $500.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Normal probability distribution
Problems of normally distributed samples can be 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:
.
What is the probability that the average amount billed on the sample bills is greater than $500?
This probability is 1 subtracted by the pvalue of Z when
. So



has a pvalue of 0.9998.
So there is a 1-0.9998 = 0.0002 = 0.02% probability that the average amount billed on the sample bills is greater than $500.