1. ## Statistics help

I cannot get this problem for the life of me. I've been working on it for at least an hour and have produced no work.

It is known that the incomes of subscribers to a particular magazine have a normal distribution with a standard deviation of $6,600. A random sample of 25 subscribers is taken. a. What is the probability that the sample standard deviation of their incomes is more than$4,000?
b. What is the probability that the sample standard deviation of their incomes is less than \$8,000?

Any help, even just a started, would be great.

2. ## Re: Statistics help

Hey blind527.

Recall the chi-square distribution where (n-1)s^2/sigma^2 has a chi-square distribution with n-1 degrees of freedom. You are given sigma^2 as 6600^2 so you can look at the distribution for s^2 by using the appropriate chi-square PDF.

Usually in statistical inference, we do the opposite: we start with s^2 and try to make inferences on sigma^2 but we can do the same thing in reverse and it is in some cases very useful to do the unconventional thing.

3. ## Re: Statistics help

Thank you for that, but I really just don't know how to apply any of this to the problem. I know that sigma^2 is 6600^2, but not sure what I do with anything relating to the problem itself.

4. ## Re: Statistics help

So since you have s^2(n-1)/sigma^2 ~ chi-square(n-1) = X^2 you want P(s^2 > 4000^2). Lets do a few transformations:

P(s^2 > 4000^2) implies
P((n-1)s^2 > 4000^2*(n-1)) which implies
P((n-1)s^2 > 4000^2*(n-1)/sigma^2) which implies
P(X^2 > 4000^2*(n-1)/sigma^2) where X^2 has a chi-square distribution with n-1 degrees of freedom.

Since P(X^2 > x) = 1 - P(X^2 < x) you can use a computer to calculate P(X^2 < x) which is just the cumulative probability for a chi-square(n-1) and x is given above in our derivation and can be calculated since we know n and sigma.

5. ## Re: Statistics help

I forgot to thank you for your help. Your explanation helped me greatly.