a) Standard Error of the mean = $980
b) Probability that the sample mean will be more than $25,050 = 0.50
c) Probability that the sample mean will be within $1,250 of the population mean = 0.7995
d) Probability that the sample mean will be within $1,250 of the population mean if the sample size is increased to 200 = 0.9692
Step-by-step explanation:
Complete 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 $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)
Solution
According to the Central limit theorem, for a normally distributed distribution where a random sample of large enough sample size with each variable independent of one another, the mean of the sampling distribution is approximately equal to the population mean and the standard deviation of the sampling distribution or the standard error of the mean (σₓ) is given as
σₓ = (σ/√n)
where σ = population standard deviation = $8,200
n = sample size = 70
Mean of sampling distribution (μₓ) = Population mean (μ) = $25,050
a) Standard Error of the mean = σₓ
= (8200/√70) = 980.09 = $980 to nearest whole number
b) Probability that the sample mean will be more than $25,050 = P(x > 25050)
We first normalize or standardize $25;050
The standardized score for any value is the value minus the mean then divided by the standard deviation.
z = (x - μₓ)/σₓ = (25050 - 25050)/980 = 0
To determine the required probability 45mg/L, P(x > 25050) = P(z > 0)
We'll use data from the normal probability table for these probabilities
P(x > 25050) = P(z > 0) = 1 - P(z ≤ 0) = 1 - 0.50 = 0.50
c) The probability that the sample mean will be within $1,250 of the population mean
(mean of sampling distribution) + 1250 = 25050 + 1250 = $26,300
(mean of sampling distribution) - 1250 = 25050 - 1250 = $23,800
Probability that the sample mean will be within $1,250 of the population mean
= P(23800 ≤ x ≤ 26300)
We first normalize $23,800 and $26,300
For $23,800
z = (x - μₓ)/σₓ = (23800 - 25050)/980 = -1.28
For $26,300
z = (x - μₓ)/σₓ = (26300 - 25050)/980 = 1.28
The required probability
P(23800 ≤ x ≤ 26300) = P(-1.28 ≤ x ≤ 1.28)
We'll use data from the normal distribution table for these probabilities
P(23800 ≤ x ≤ 26300) = P(-1.28 ≤ x ≤ 1.28)
= P(z ≤ 1.28) - P(z ≤ -1.28)
= 0.89973 - 0.10027
= 0.79946 = 0.7995 to 4 d.p
d) How would the probability in part (c) change if the sample size were increased to 200?
If the sample size increase to 200, the standard deviation of the sampling distribution or standard error of the mean changes to
σₓ = (σ/√n)
where σ = population standard deviation = $8,200
n = sample size = 200
Standard Error of the mean = σₓ
= (8200/√200) = 579.83 = $580 to nearest whole number
Probability that the sample mean will be within $1,250 of the population mean
= P(23800 ≤ x ≤ 26300)
We first normalize $23,800 and $26,300
For $23,800
z = (x - μₓ)/σₓ = (23800 - 25050)/580 = -2.16
For $26,300
z = (x - μₓ)/σₓ = (26300 - 25050)/980 = 2.16
The required probability
P(23800 ≤ x ≤ 26300) = P(-2.16 ≤ x ≤ 2.16)
We'll use data from the normal distribution table for these probabilities
P(23800 ≤ x ≤ 26300) = P(-2.16 ≤ x ≤ 2.16)
= P(z ≤ 2.16) - P(z ≤ -2.16)
= 0.98461 - 0.01539
= 0.96922 = 0.9692 to 4 d.p
Hope this Helps!!!
