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)

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

answer: 0,6

step-by-step explanation:

this would make it the y intercept.