Question

The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): critical reading-502; mathematics-515; writing-494; Assume that the population standard deviation on each part of the test is = 100.
What is the probability a random 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)?

Answer 1

Are we assuming normal distribution? If so, then calculate
\[z = \frac{ xbar - \mu }{ \sigma/\sqrt{n} }\] using the known values for mu and sigma: Xbar is the sample mean from the 90 students.
We want to know probability of being within 10 points of mu = 515 on Math score, so within 505 and 525.
P(505 < Xbar < 525) now calculate z for 505 and 525:
For Xbar = 505,
z = \[\frac{ 505-515 }{ 100/\sqrt{90} } = -0.95\]
For Xbar = 525,
z = \[\frac{ 525-515 }{ 100/\sqrt{90} } = 0.95\]

Now we need P(-0.95 < Z < 0.95)
You can use a z-table or use Excel or some other statistics program. From the z-table,
P(z<.95) = 0.8289, but this includes everything from 0.95 down to the very bottom of the normal distribution. We want the probability between Z = 0 and Z = 0.95, so we can take the table value and subtract 0.5 from it. 0.8289 - 0.5 = 0.3289

Because of symmetry, we know that the probability between z=-0.95 and z=0 is also 0.3289, so the probability of the sample mean from the sample of 90 students being between 505 and 525 is the same as the probability of z being between -0.95 and 0.95, which is 0.3289 + 0.3289 = .6578 or 65.78%. This shouldn't be too surprising since this is really close to the probability of being within 1 standard deviation of the mean, which we know is ~ 68%. For these students, instead of 1 standard deviation, the question is +/- 0.95 standard deviations, almost 1. So the answer should be close to 68%. You can state that the probability of the sample of 90 students scoring, on average, within 10 points of 515 is 65.78%.

Answer 2

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