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Mathematics 14 Online
OpenStudy (destinyyyy):

Looking for help with three elementary statistics problems..

OpenStudy (destinyyyy):

The height of women are normally distributed with a mean of 65 inches and a standard deviation of 2.5 inches. The heights of men are also normal with a mean of 70 inches. approximately what percent of women are taller than a man of average height? I got to 0.0228

OpenStudy (error1603):

k

OpenStudy (mhchen):

We start by finding z-score of men's average height in the women's graph. Z - score = (70-65)/2.5 = 2 Then we use a calculator or Table A to find the percentile: 97.725% That's the number of women below men's height. To find the number of women above men's height: 100% - 97.725% = 0.0228ish So yes, you are correct. I have a statistics test next period so thanks for helping me study as well :)

OpenStudy (error1603):

Copied.

OpenStudy (destinyyyy):

1- 0,97725= 0.02275= 0.0228 not "100% - 97.725% = 0.0228ish " Also I know a step is missing because that is not listed as an option. It asks for the percent.

OpenStudy (destinyyyy):

0.97725 *

OpenStudy (destinyyyy):

Why are you saying "k" and "copied"?

OpenStudy (holsteremission):

Given any decimal, you can convert to a percentage by multiplying by \(100\). Example: \(0.05\times100=5\%\).

OpenStudy (destinyyyy):

Alright. I guess ill go with 2%.

OpenStudy (destinyyyy):

OpenStudy (holsteremission):

By comparison, the original distribution has a larger standard deviation than any one of the (larger) sampling distributions. So as the sample size \(n\) grows, the distribution get narrower, meaning the standard deviation get smaller, which in turn means the same thing as "less variable".

OpenStudy (destinyyyy):

So the answer is A. I thought it might be

OpenStudy (holsteremission):

Yup

OpenStudy (destinyyyy):

Can you help with one more?

OpenStudy (holsteremission):

Sure

OpenStudy (destinyyyy):

Okay. Thank you. If the standard deviation of a population is 20 and we take a sample of size 16 from which to calculate a mean, the standard error (the standard deviation of the sample mean) is:

OpenStudy (holsteremission):

I believe the standard error is given by \(\dfrac{\sigma}{\sqrt n}\) (standard deviation divided by square root of sample size).

OpenStudy (destinyyyy):

Alright. So the answer is 5

OpenStudy (holsteremission):

Yes

OpenStudy (destinyyyy):

Thanks for the help.

OpenStudy (holsteremission):

yw

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