Given that 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
σ =
100.
We 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 (to 4 decimals) as follows:
Within 10 points of 502 implies (502 - 10, 502 + 10) = (492, 512).
Because the sample is large, we assume normal distribution.
The probability that the mean statistic of sample data is between two points (a, b) is given by:
[tex]P(a\ \textless \ \bar{x}\ \textless \ b)=P\left( \frac{b-\mu}{\sigma/\sqrt{n}} \right)-P\left( \frac{a-\mu}{\sigma/\sqrt{n}} \right)[/tex]
Thus,
[tex]P(492\ \textless \ \bar{x}\ \textless \ 512)=P\left( \frac{512-502}{100/\sqrt{90}} \right)-P\left( \frac{492-512}{100/\sqrt{90}} \right) \\ \\ = P\left(\frac{10}{10.5409} \right)-P\left(\frac{-10}{10.5409} \right)=P(0.9487)-P(-0.9487) \\ \\ =P(0.9487)-[1-P(0.9487)]=2P(0.9487)-1=2(0.82861)-1 \\ \\ =1.65722-1=0.65722[/tex]
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 (to 4 decimals) is 0.6572.
