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October 31, 2014

October 31, 2014

Posted by **Anonymous** on Sunday, October 17, 2010 at 8:21pm.

Assume that the mean systolic blood pressure of normal adults is 120 millimeters of mercury and the standard deviation is 5.6. Assume the variable is normally distributed:

A)If an individual is selected, find the probability that the individual;s pressure will be between 120 and 121.8

B) If a sample of 30 adults is randomly selected, find the probability that the sample mean will be between 120 and 121.8

At a large publishing company, the mean age of proofreaders is 36.2 years, and the standard deviation is 3.7 years. Assume the variable is normally distributed:

A) If a proofreader from the company is randomly selected, find the probability that his or her age will be between 36 and 37.5 years.

B) If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders will be between 36 and 37.5 years

- Statistics -
**PsyDAG**, Monday, October 18, 2010 at 2:30pmZ = (Sample mean-population mean)/SEm = (38.2-37)/SEm

SEm (Standard Error of the Mean) = SD/√(n-1)

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score.

Use similar process for B sections of other questions

A) Z = (score-mean)/SD = (121.8-120)/5.6

Use the same table and similar process for both A questions.

The difference between A and B is that the former deals with a distribution of scores and the latter deals with a distribution of means.

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