Question

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?

Answer 1

Answer:

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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

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:

$\mu = 10348, \sigma = 2500, n = 100, s = \frac{2500}{\sqrt{100}} = 250$

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

$Z = \frac{100}{250} = 0.4$

$Z = -\frac{100}{250} = -0.4$

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

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{12600 - 10348}{250}$

$Z = 9$

$Z = 9$ has a pvalue of 1.

1 - 1 = 0

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