1. Say age or weight used as a variable is distributed evenly over a
big group of people. A standard deviation of ? and a mean of µ.

You collect many samples of size n = 100 and find the mean of the
variable for each sample. How does the standard deviation of the
distribution of sample means compare to ?? Explain.

2. It's a race that has a mean of 61 minutes and a standard deviation
of 9 minutes with normal distribution. What % of people finishing
times are less than 55 minutes? With a random sample of 25 runners
selected and the mean finishing time is computed. What % of the sample
means are less than 55 minutes?

3. What % of peoples finishing times are between 60 & 62 minutes? With
a random sample of 36 runners and the mean finishing time of the
sample is computed. What % of sample means are between 60 & 62
minutes?

Hi jeffstud!
Here are the answers to your questions.

Question 1

At the following link (under part 2) you'll find how the mean and
variance of the sample mean relates to the mean and variance of the
population.

Distribution of the Sample Mean
http://www.ms.uky.edu/~viele/sta417s00/sampdist/sampdist.html

Specifically, we have that, if the population variance is ?^2, then
the variance of the sample mean is ?^2/n. Therefore, the standard
deviation of the distribution of sample means is simply ?/sqrt(n)
[sqrt() means square root of]. In this case, with n=100,

?/sqrt(100) = ?/10

Therefore, the standard deviation of the sample mean is ten times
smaller than the standard deviation of the population. As the sample
size grows, this std. dev. gets smaller.


Question 2

Calling X the time needed to finish the race, we have that X~N(61,9).
We want to find here

Prob(X < 55)

In order to find this probability, we need to use a standard normal
distribution table. You can find one in any Statistics book or at the
following link

z-distribution
http://www.math2.org/math/stat/distributions/z-dist.htm

Since the table gives the probabilities for the standard normal
distribution, we must first "convert" X to a standard normal
distribution by subtracting its mean and dividing it by its standard
deviation. So we have that:

 Prob( X < 55 )
=Prob( (X-61)/9 < (55-61)/9 )
=Prob(     Z    <   -0.666  )

I renamed (X-61)/9 as Z because the variable "(X-61)/9" follows a
standard normal distribution (that is, a normal distribution with mean
0 and variance 1).

Now we look up the value -0.666 in the table, getting that:

Prob(Z < -0.666) = 0.254

Therefore, 25.4% of the people finish the race in less than 55 minutes.

In order to answer the second part of this question, we make use of
the results we found in the question 1. Calling Xbar to the sample
mean, we know that Xbar follows a normal distribution with mean 61 and
standard deviation 9/sqrt(25)=1.8. So Xbar~N(61,1.8). We solve this in
the exact same way as before:

 Prob(Xbar < 55)
=Prob( (Xbar-61)/1.8 < (55-61)/1.8 )
=Prob(Z < -3.33)
=0.00043

Therefore, 0.043% of the sample means are less than 55 minutes.


Question 3

This question is solved in the same way as before:

 Prob(60 < X < 62)
=Prob(X<62) - Prob(X<60)
= (subtract the mean and divide by std. dev. ...)
=0.5442 - 0.4557
=0.0884

So 8.84% of the people finish in between 60 and 62 minutes.

We're also asked to find:

 Prob(60 < Xbar < 62)
=Prob(Xbar<62) - Prob(Xbar<60)
= (subtract the mean and divide by std. dev. ...)
=0.4214

Thus, 42.14% of sample means fall between 60 and 62.


Google search terms
"normal distribution" table
://www.google.com/search?hl=en&q=%22normal+distribution%22+table&btnG=Google+Search
"sample mean" properties

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Clarification of Answer by elmarto-ga on 06 Apr 2005 11:12 PDT

I accidentally pressed the "Answer Question" button before finishing.

Google search terms
"normal distribution" table
://www.google.com/search?hl=en&q=%22normal+distribution%22+table&btnG=Google+Search
"sample mean" properties
://www.google.com/search?hl=en&lr=&q=%22sample+mean%22+properties


I hope this helps! If you have any questions regarding my answer,
please don't hesitate to request a clarification. Otherwise I await
your rating and final comments.

Best wishes!
elmarto

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Request for Answer Clarification by jeffstud-ga on 06 Apr 2005 12:55 PDT

Hi, thanks for the quick reply, I was wondering if I could just get
the answers without having to visit any reference sites. Thanks for
the reference sites though, please let me know,

Jeff.

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Clarification of Answer by elmarto-ga on 06 Apr 2005 15:01 PDT

Hi jeffstud,
The answers are right there, without you having to visit any reference
link. The first link is just in case you want more information
regarding properties of the sample mean. The second link is a table of
numbers you can find in any Statistics book, I just placed it in case
you don't have easy access to one.

Best wishes,
elmarto

Great answers, quick response as well