Question

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 $26, 800. Assume the population standard deviation is $6,600 . Suppose that a random sample of 50 USC students will be taken from this population.

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 $26,800? (to 2 decimals)
c. What is the probability that the sample mean will be within $500of the population mean? (to 4 decimals)
d. How would the probability in part (c) change if the sample size were increased to $160? (to 4 decimals)

Answer 1

Answer:

Part a: The standard error of the mean is $933.

Part b: The probability that the sample mean will be more 26800 is 0.5

Part c: The probability that the sample mean will be within 500 is 0.4078

Part d: The probability that the sample mean will be within 500 for n=160 is 0.6620. The probability is increased by 0.2542.

Step-by-step explanation:

Part a:

The value of standard error is given as

$\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$

Now the value of standard deviation and number of samples is given as

σ=$6600

n=50

$\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}\\\sigma_{\bar{X}}=\frac{6600}{\sqrt{50}}\\\sigma_{\bar{X}}=\ 933$

So the standard error of the mean is $933.

Part b:

X=$26800

So P(X>26800)=1-P(X≤26800)

For this value the z square is given as

$z=\frac{(X-\mu)\sigma}{\sqrt{n}}\\z=\frac{(26800-26800)6600}{\sqrt{50}}\\z=0$

P(X≤6800)=P(z≤0)

From the z table the value is

P(z≤0)=0.5

P(X>26800)=1-P(X≤26800)

P(X>26800)=1-0.5

P(X>26800)=0.5

So the probability that the sample mean will be more 26800 is 0.5

Part c:

The values remain within the $500 of the mean are X>26300 and X<27300.

So the Probability is given as

P(26300≤X≤27300)=P(X≤27300)-P(X≤26300)

For the values of X the z score is given as

$z1=\frac{(X1-\mu)\sigma}{\sqrt{n}}\\z1=\frac{(26300-26800)6600}{\sqrt{50}}\\z1=-0.54$

$z2=\frac{(X2-\mu)\sigma}{\sqrt{n}}\\z2=\frac{(27300-26800)6600}{\sqrt{50}}\\z2=0.54$

P(X≤27300)-P(X≤26300)=P(Z≤Z2)-P(Z≤Z1)

                                        =P(Z≤0.54)-P(Z≤-0.54)

                                        =0.7039-0.2961

                                       = 0.4078

So the probability that the sample mean will be within 500 is 0.4078

Part d:

The values remain within the $500 of the mean are X>26300 and X<27300 and n=160 so.

So the Probability is given as

P(26300≤X≤27300)=P(X≤27300)-P(X≤26300)

For the values of X the z score is given as

$z1=\frac{(X1-\mu)\sigma}{\sqrt{n}}\\z1=\frac{(26300-26800)6600}{\sqrt{160}}\\z1=-0.96$

$z2=\frac{(X2-\mu)\sigma}{\sqrt{n}}\\z2=\frac{(27300-26800)6600}{\sqrt{160}}\\z2=0.96$

P(X≤27300)-P(X≤26300)=P(Z≤Z2)-P(Z≤Z1)

                                        =P(Z≤0.96)-P(Z≤-0.96)

                                        =0.8310-0.1690

                                       = 0.6620

So the probability that the sample mean will be within 500 for n=160 is 0.6620. The probability is increased by 0.2542.