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October 23, 2014

October 23, 2014

Posted by **cindy** on Monday, July 19, 2010 at 10:45am.

A. What is the likelihood that an individual woman has a blood pressure abouve 125?

B. Suppose a random sample of n=300 women is selected and the mean blood pressure for tyhe sample is computed. What is the likelihood that the mean blood pressure for the sample will be above 125?

C. Suppose a random sample of n=300 women is selected and the mean blood pressure for the sample is computed. What is the likelihood that the mean blood pressure for the sample will be bleow. 114?

- statistical reasoning -
**PsyDAG**, Monday, July 19, 2010 at 1:13pmA. Z = (score-mean)/SD (for distribution of scores)

B & C. Z = (mean1 - mean2)/standard error (SE) of difference between means (for distribution of means)

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√(n-1)

Since only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to these Z scores.

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