Suppose data made available through a health system tracker showed health expenditures were $10,348 per person in the United States. Use $10,348 as the population mean and suppose a survey research firm will take a sample of 100 people to investigate the nature of their health expenditures. Assume the population standard deviation is $2,500. Required:
a. What is the probability the sample mean will be within ±$100 of the population mean?
b. What is the probability the sample mean will be greater than $12,600?

a) 30.08% probability the sample mean will be within $100 of the population mean.

b) 0% probability the sample mean will be greater than $12,600

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:

a. What is the probability the sample mean will be within ±$100 of the population mean?

This is the pvalue of Z when X = 100 divided by s subtracted by the pvalue of Z when X = -100 divided by s. So

Z = 0.4 has a pvalue of 0.6554, Z = -0.4 has a pvalue of 0.3556

0.6554 - 0.3546 = 0.3008

30.08% probability the sample mean will be within $100 of the population mean.

b. What is the probability the sample mean will be greater than $12,600?

This is 1 subtracted by the pvalue of Z when X = 12600. So

By the Central Limit Theorem

has a pvalue of 1.

1 - 1 = 0

0% probability the sample mean will be greater than $12,600

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step-by-step explanation: