The latest available data showed health expenditures were $8,086 per person in the United States or 17.6% of Gross Domestic Product (Centers for Medicare & Medicaid Services website, April 1, 2012). Use $8086 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 $2500.Show the sampling distribution of the mean amount of health care expenditures for a sample of 100 people. Round your answer to nearest whole value.What is the probability the sample mean will be within � $200 of the population mean? Round your answer to four decimal placesWhat is the probability the sample mean will be greater than $9000? Round your answer to four decimal places

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24-09-2019
Mathematics

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The latest available data showed health expenditures were $8,086 per person in the United States or 17.6% of Gross Domestic Product (Centers for Medicare & Medicaid Services website, April 1, 2012). Use $8086 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 $2500.Show the sampling distribution of the mean amount of health care expenditures for a sample of 100 people. Round your answer to nearest whole value.What is the probability the sample mean will be within � $200 of the population mean? Round your answer to four decimal placesWhat is the probability the sample mean will be greater than $9000? Round your answer to four decimal places

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joaobezerra joaobezerra · Answer 1 · 2019-09-26T22:35:07+02:00

Answer:

There is a 6.378% probability that the will be within $200 of the population mean.

There is a 35.569% probability that the sample mean will be greater than $9000.

Step-by-step explanation:

Normal model problems can be solved by the zscore formula.

On a normaly distributed set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a value X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

Each z-score value has an equivalent p-value, that represents the percentile that the value X is, or the probability that a value is LOWER than the value of X.

In our problem, we have these following informations:

The population mean is $8086. So [tex]\mu = 8086[/tex].

The population standard deviation is $2500. So [tex]\sigma = 2500[/tex]

What is the probability the sample mean will be within $200 of the population mean?

Within 200 from the the population mean is from 8086-200 through 8086+200. So from 7886 through 8286. The answer to this question is the pvalue of the zscore of [tex]X = 8286[/tex] subtracted by the pvalue of the zscore of [tex]X = 7886[/tex].

We find the pvalue of a zscore looking at the zscore table.

For X = 8286

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{8282-8086}{2500}[/tex]

[tex]Z = 0.08[/tex]

The pvalue of [tex]Z = 0.08[/tex] is .53188.

For X = 7886

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{7886-8086}{2500}[/tex]

[tex]Z = -0.08[/tex]

The pvalue of [tex]Z = -0.08[/tex] is .4681.

The probability that the sample mean will be within $200 of the population mean is

[tex].53188 - .4681 = 0.06378[/tex]

There is a 6.378% probability that the will be within $200 of the population mean.

What is the probability the sample mean will be greater than $9000?

The first step to find this probability is finding the Z-score of [tex]X = 9000[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{9000-8086}{2500}[/tex]

[tex]Z = 0.37[/tex]

The pvalue of [tex]Z = 0.37[/tex] is .64431. This means that there is a 64.431% probability that the sample mean will be lesser than $9000.

The probability that the sample mean is:

100% - 64.431% = 35.569%

There is a 35.569% probability that the sample mean will be greater than $9000.