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

To calculate the probability that the sample mean will be within ±$100 of the population mean, we can use the Central Limit Theorem. The probability is approximately 0.6554.

Step-by-step explanation:

To calculate the probability that the sample mean will be within ±$100 of the population mean, we can use the Central Limit Theorem. Since the sample size is large (n > 30), the sampling distribution of the sample mean will be approximately normal. The standard deviation of the sampling distribution, also known as the standard error, can be calculated by dividing the population standard deviation by the square root of the sample size.

Standard error = population standard deviation / sqrt(sample size) = $2,500 / sqrt(100) = $250

Now we can calculate the z-score for the ±$100 range. The z-score for ±$100 means that we want to find the area under the normal curve within ±$100 from the mean. To do this, we can convert the dollar amount to a z-score using the formula: z = (x - mean) / standard error

For +$100:

z = ($10,348 + $100 - $10,348) / $250 = 0.4

Using the z-table or a calculator, we can find that the area to the left of 0.4 is 0.6554. Therefore, the probability that the sample mean will be within +$100 of the population mean is approximately 0.6554.

For -$100:

z = ($10,348 - $100 - $10,348) / $250 = -0.4

The area to the left of -0.4 is the same as the area to the right of 0.4. Therefore, the probability that the sample mean will be within -$100 of the population mean is also approximately 0.6554.

Answer 2

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 = (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 = (\sigma)/(√(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 = (2500)/(√(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 = (100)/(250) = 0.4`

`Z = -(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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