Posted by **Diandra** on Tuesday, October 26, 2010 at 9:46am.

find the 90% confidence interval for the variance and standard deviation of the ages of seniors at Oak Park College if a sample of 24 students has a standard deviation of 2.3 years. Assume the variable is normally distributed. Can someone explain the steps to solve this? Thanks so much.

- Statistics -
**PsyDAG**, Tuesday, October 26, 2010 at 10:32am
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the Z score for .4500 from mean (Z = 1.645)

90% conf. Interval = mean ± 1.645(SD)

- Statistics -
**mona**, Tuesday, October 26, 2010 at 9:12pm
A party host gives a door prize to one guest chosen at random. There are 48 men and 42 women at the party. What is the probability that the prize goes to a woman? Round your answer to 3 decimal places.

Answer

- Statistics -
**Bluegravity**, Wednesday, December 12, 2012 at 1:52pm
That is incorrect. You do not use the Z-score for the mean. This is a Chi-square distribution.

1-.9=.10/2 (for left and right alpha tail) = .05

Find .05 and .95 using the d.f.=23

That equals 34.172 and 13.091

Using the formula for the variance which is ((n-1)s^(2))/(x^(2))

Your equation is an interval, so if you plug each in, your answer should be:

3.46<variance<9.29

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