The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009):
Assume that the population standard deviation on each part of the test is = 100.
a. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test (to 4 decimals)?
b. What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test (to 4 decimals)?
c. What is the probability a sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test (to 4 decimals)?
SEm = SD/√n
10 = ± Z (SEm)
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
To solve these probability problems, we will use the concept of sampling distributions and the normal distribution.
a. To find the probability that a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test, we need to calculate the z-score.
The formula for calculating the z-score is:
Z = (X - μ) / (σ / √n)
Where:
X = sample mean
μ = population mean
σ = population standard deviation
n = sample size
In this case,
X = population mean = 502
μ = population mean = 502
σ = population standard deviation = 100
n = sample size = 90
Calculating the z-score:
Z = (X - μ) / (σ / √n)
= (502 - 502) / (100 / √90)
= 0 / (100 / √90)
= 0
The z-score is 0.
Now, we want to find the probability that the sample mean test scores are within 10 points of the population mean. This is equivalent to finding the probability that the z-score is between -10/100 and 10/100, which can be calculated using a standard normal distribution table or a calculator.
Using a standard normal distribution table, the probability corresponding to a z-score of -10/100 is approximately 0.4602, and the probability corresponding to a z-score of 10/100 is approximately 0.5398.
Therefore, the probability that a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test is:
0.5398 - 0.4602 = 0.0796 (rounded to 4 decimals)
b. We can follow the same steps for the Mathematics part of the test.
X = sample mean = 515
μ = population mean = 515
σ = population standard deviation = 100
n = sample size = 90
Calculating the z-score:
Z = (X - μ) / (σ / √n)
= (515 - 515) / (100 / √90)
= 0 / (100 / √90)
= 0
Using a standard normal distribution table, the probability corresponding to a z-score of -10/100 is approximately 0.4602, and the probability corresponding to a z-score of 10/100 is approximately 0.5398.
Therefore, the probability that a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test is:
0.5398 - 0.4602 = 0.0796 (rounded to 4 decimals)
c. We can follow the same steps for the Writing part of the test.
X = sample mean = 494
μ = population mean = 494
σ = population standard deviation = 100
n = sample size = 100
Calculating the z-score:
Z = (X - μ) / (σ / √n)
= (494 - 494) / (100 / √100)
= 0
Using a standard normal distribution table, the probability corresponding to a z-score of -10/100 is approximately 0.4602, and the probability corresponding to a z-score of 10/100 is approximately 0.5398.
Therefore, the probability that a sample of 100 test takers will provide a sample mean test score within 10 points of the population mean of 494 on the Writing part of the test is:
0.5398 - 0.4602 = 0.0796 (rounded to 4 decimals)