I'm having trouble with some questions.
Problem 1.
If we have a continious random variable X that is uniformly distributed on the
interval (0,15).

Find the mean, variance, first quartile.
(Note) A property of a uniform random variable defined on the interval (a,b) is that the variance is
1/12(b-a)^2.


Problem 2.
IF we have a random sample (with sample size n=10) from a population with a uniform distribution on the interval (0,15).
a)Find the mean, variance, standard of error of the sampling distribution of the sample mean.

b) Now use the CENTRAL LIMIT THEOREM. Find the first quartile.



I would appreciate your help. Just how to start will be helpful.

Problem 1.

The mean will be the half-way mark between 0 and 15.
The variance will be (1/12)*(15-0)^2.
The first quartile will be 25% of the way between 0 and 15 (i.e., divide the mean by 2).

Problem 2.

The sample mean will be the same as the mean in problem 1.
The sample variance, over the long run, will be the same as the variance in Problem 1.
The standard error of the sample mean will be:
sqrt(variance)/n

Using the C.L.T., the first quartile will be where z = -0.674
-0.674 = (x - mu) / (s / sqrt(n))

You're given that n=10, so plug that in, and s= sqrt(variance), and use your answers from the previous work to solve for x.

Post back if you get stuck....

Hi Johnn,
wow.. Thanks!! I'll try to solve it now..

I was wondering for the first question.
If we were asked to find the third quartile....
What will be dividing the mean by? (3/4)?
And how about the second quesiton?

Edit.
For problem 2, wouldn't the mean be the answer from problem 1 divided by 10?
so, 7.5/10.

Michelle.:wave:

Problem 2.
Using the C.L.T., the first quartile will be where z = -0.674
-0.674 = (x - mu) / (s / sqrt(n))


I'm stuck with here.. I don't know how the z was obtained.

Michelle,

The third quartile would be (3/4)*15

If you had a continuous random variable from 0 to 20, the mean would be 10, the first quartile would be 5, and the third quartile would be 15.

In question 2b) I think they're asking for the first quartile of the sampling distribution of means, not the population distribution.

For problem 2, wouldn't the mean be the answer from problem 1 divided by 10?
so, 7.5/10.
- the mean of the sampling distribution (of means) equals the mean of the population.

JohnM

Problem 2.
Using the C.L.T., the first quartile will be where z = -0.674
-0.674 = (x - mu) / (s / sqrt(n))


I'm stuck with here.. I don't know how the z was obtained.


Using the normal distribution tables - find the value of z that has 25% of the area (under the curve) below it

Using the normal distribution tables - find the value of z that has 25% of the area (under the curve) below it


Got it!!
why is it a negative value??


--------------*-----
0 z

Got it!!
why is it a negative value??


--------------*-----
0 z

The standard normal curve is centered at 0, and z=0 marks the spot where 50% of the area is below it. If only 25% of the area is below a certain value of z, it must be less than 0.

Attached is an image that shows the "green" area as the lower 25% of the normal curve, and the dividing line is below 0.