Answer:
0.2206 = 22.06% probability that the sample mean is greater than 60.
Explanation:
To solve this question, we have to understand the normal probability distribution and the Central Limit Theorem.
Normal probability distribution:
Problems of normally distributed samples are 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.
Central Limit Theorem:
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

In this problem, we have that:

Compute the probability the sample mean is greater than 60:
This probability is 1 subtracted by the pvalue of Z when X = 60. So

By the Central Limit Theorem



has a pvalue of 0.7794
1 - 0.7794 = 0.2206
0.2206 = 22.06% probability that the sample mean is greater than 60.