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After deducting grants based on need, the average cost to attend the University of Southern California (USC) is $27,175 (U.S. News & World Report, America’s Best Colleges, 2009 ed.). Assume the population standard deviation is $7,400. Suppose that a random sample of 60 USC students will be taken from this population.

a. Refer to Exhibit 2. What is the value of the standard deviation of the mean? (Note: keep two decimal places.)

b.What is the probability that the sample mean will be more than $27,175

c.What is the probability that the sample mean will be within $1,000 of population mean? (Note: keep two decimal places for the z value and four decimal places for the final probability value.)

d.What would be the probability in Part c if the sample size were increased to 100? (Note: keep two decimal places for the z value and four decimal places for the final probability value.)

Respuesta :

Using the normal distribution and the central limit theorem, it is found that:

a) The standard deviation is of $955.34.

b) 0.5 = 50% probability that the sample mean will be more than $27,175.

c) There is a 0.2938 = 29.38% probability that the sample mean will be within $1,000 of population mean.

d) There is a 0.177 = 17.7% probability that the sample mean will be within $1,000 of population mean.

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].

In this problem:

  • The average is of $27,175, hence [tex]\mu = 27175[/tex].
  • The population standard deviation is of $7,400, hence [tex]\sigma = 7400[/tex].
  • A sample of 60 students is taken, hence [tex]n = 60[/tex].

Item a:

[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{7400}{\sqrt{60}} = 955.34[/tex]

The standard deviation is of $955.34.

Item b:

This probability is 1 subtracted by the p-value of Z when X = 27175, hence:

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

By the Central Limit Theorem

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

[tex]Z = \frac{27175 - 27175}{0}[/tex]

[tex]Z = 0[/tex]

[tex]Z = 0[/tex] has a p-value of 0.5.

1 - 0.5 = 0.5

0.5 = 50% probability that the sample mean will be more than $27,175.

Item c:

[tex]Z = \frac{1000}{s}[/tex]

[tex]Z = \frac{1000}{955.34}[/tex]

[tex]Z = 1.05[/tex]

The probability is P(|Z| > 1.05), which is 2 multiplied by the p-value of Z = -1.05.

Looking at the z-table, Z = -1.05 has a p-value of 0.1469.

2 x 0.1469 = 0.2938

There is a 0.2938 = 29.38% probability that the sample mean will be within $1,000 of population mean.

Item d:

Sample of 100, hence [tex]n = 100, s = \frac{7400}{\sqrt{100}} = 740[/tex]

[tex]Z = \frac{1000}{s}[/tex]

[tex]Z = \frac{1000}{740}[/tex]

[tex]Z = 1.35[/tex]

Z = -1.35 has a p-value of 0.0885.

2 x 0.0885 = 0.177

There is a 0.177 = 17.7% probability that the sample mean will be within $1,000 of population mean.

To learn more about the normal distribution and the central limit theorem, you can take a look at https://brainly.com/question/24663213

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