After deducting grants based on need, the average cost to attend the University of Southern California (USC) is $27,175. Assume the population standard deviation is $7,400. Suppose that a random sample of 70 USC students will be taken from this population.

(a)

What is the value of the standard error of the mean? (Round your answer to the nearest whole number.)

$  

(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 the population mean? (Round your answer to four decimal places.)

 

(d)

What is the probability that the sample mean will be within $1,000 of the population mean if the sample size were increased to 100? (Round your answer to four decimal places.)

The value of the standard error of the mean =884.

Step-by-step explanation

a) The value of the standard error of the mean :

a) Std deviation of the mean =population std deviation/sqrt(n)=σ/√n =7400/√70=884.46 = 884(approximately).

Hence, the value of the standard error of the mean =884.

b) The probability that the sample mean will be more than $27,175 :

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

c) Probability that the sample mean will be within $1,000 of population mean:

P(|x -μ|<1000)=P(-1000/884<Z<1000/884=P(-1.13<Z<1.13)= 1-2p(z<-1.13)=1- 2*0.1292=0.7416

Hence, the probability that the sample mean will be within $1,000 of the population mean =0.7416.

d)  The probability that the sample mean will be within $1,000 of the population mean if the sample size were increased to 100 :

std deviation of the mean =population std deviation/sqrt(n)=7400/sqrt(100)=740

probability that the sample mean will be within $1,000 of population mean

=P(-1000/740<Z<1000/740)=P(-1.35<Z<1.35)= 1-2p(z<-1.35)=1- 2*0.0885=0.8230

Hence, probability that the sample mean will be within $1,000 of population mean if n=100 is 0.8230.

Please see the attached file for the complete solution