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

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 and standard deviation , the z-score of a value X is given by:

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 .

The population standard deviation is $2500. So

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 subtracted by the pvalue of the zscore of .

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

For X = 8286

The pvalue of is .53188.

For X = 7886

The pvalue of is .4681.

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

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

The pvalue of 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.

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17