Medicare Hospital Insurance The average yearly Medicare Hospital Insurance benefit per person was $4064 in a recent year. Suppose the benefits are normally distributed with a standard deviation of $460. Assume that the sample is taken from a large population and the correction factor can be ignored. Use a TI-83 Plus/TI-84 Plus calculator. Round your answer to at least four decimal places. Find the probability that the mean benefit for a random sample of 20 patients is more than $4100.
P (X > 4100) =?

0.3632 = 36.32% probability that the mean benefit for a random sample of 20 patients is more than $4100.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use 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 normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

Population:

Sample of 20:

Find the probability that the mean benefit for a random sample of 20 patients is more than $4100.

This is 1 subtracted by the pvalue of Z when X = 4100. So

By the Central Limit Theorem

has a pvalue of 0.6368

1 - 0.6368 = 0.3632

0.3632 = 36.32% probability that the mean benefit for a random sample of 20 patients is more than $4100.