I will start by rephrasing your question a bit:
36 samples are independently chosen from a normal distribution with
mean 1200 and
standard deviation 60. Let M be the sample mean of these samples.
What is:
a) P(M>1224)
b) P(M<1230)
c) P(1200<= M <= 1214)
This looks like almost the same question as your SAT one, but of
course real SATs cannot be normally distributed (for example, negative
scores are impossible, as are scores over 1600) and we are making
explicit that the sample is selected *independently* from the same
distribution.
Now, we can think of M itself as being a random variable.
Although it's easy to derive independently, I think here it might just
be worth knowing this theorem:
"Sampling distribution of the mean
The sampling distribution of the sample mean from a random sample of
size n drawn from a population
with mean mu and variance v will have mean mu and variance v/n ."
[Reference: Statistical Methods p. 90]
Finally, it is a theorem, and it is also intuitively clear, that the
sample mean of a normal popluation is
itself normal.
Since the standard deviation of the original distribution is 60, the
variance is 60*60=3600 .
Hence, M is a normal distribution with mean 1200 and variance 3600/36
= 100, so the standard deviation of M
is sqrt(100)=10.
Now all we need, is, given a normal distribution of mean 1200 and
standard deviation 10, what is the probability
that this distribution will lie within some range?
Generally there are two ways to solve this kind of problem. One way is
just to ask your favorite statistics software the answer directly. The
other way is to transform the problem into a question about normal
distributions with mean 1 and standard deviation 1 and then ask your
favorite statistics software (or use a table).
Since the commenter gmac-ga already posted the very helpful link to a
normal distribution calculator, I'll use that link:
http://psych.colorado.edu/~mcclella/java/normal/accurateNormal.html
I went to that link and entered in our sample mean, 1200, in the Mean
box.
Then I entered in our sample standard deviation, 10, in the StDev box.
Finally I set start and end to the appropriate values (using 10000000
as a very large value).
Here are the three answers I get:
a) start: 1224 end: 1000000:
probability is 0.082
b) start: -100000 end: 1230:
probability is 0.9987
c) start: 1200 end: 1214
probability is 0.4192
This normal calculator also has some pretty pictures (well, pictures
anyway) of the normal distribution; it's probably useful to play
around with it a bit if you want to get some intuition about normal
distributions.
I hope this is helpful. If you have any other questions, please feel
free to ask for clarification. |
