According to U.S. News & World Report's publication America's Best Colleges, the average cost to attend the University of Southern California (USC) after deducting grants based on need is $25,050. Assume the population standard deviation is $8,200 . Suppose that a random sample of 70 USC students will be taken from this population. Use z-table.a. What is the value of the standard error of the mean? (to nearest whole number)b. What is the probability that the sample mean will be more than $25,050 ? (to 2 decimals)c. What is the probability that the sample mean will be within $1,250 of the population mean? (to 4 decimals)d. How would the probability in part (c) change if the sample size were increased to 200 ? (to 4 decimals)

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MrSmoot MrSmoot · Answer 1 · 2019-04-24T01:41:02+02:00

Answer:

a: 980

b: 0.500

c: 0.7994

d: 0.9692

Step-by-step explanation:

For the first situation, we have:

µ = 25,050

σ = 8,200

n = 70

For a: The standard error is the population standard deviation divided by the square root of the sample size, so

E = 8,200/√70 = 980 (rounded to the nearest whole number from 980.09)

for b: P(x > 25,050)

Find the z-score for x > 25,050, this will be the area under the curve to the right of 25,050, or

1 - P(x < 25,050). Find P(x < 25,050)

First find the z-score for this: z = (25,050 - 25,050)/(8,200/√70) = 0.00

P(z < 0.00) = 0.500

So P(x > 25,050) = 1 - 0.500 = 0.500

For c: The probability of the sample mean being withing 1,250 gives us a range for the sample mean of 23,800 - 26,300.

See attached photo 1 for the calculations of these z-scores

For d: We do the same as part c, just change the sample size to 200. See attached photo 2 for these calculations