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October 25, 2014

October 25, 2014

Posted by **ME** on Saturday, November 17, 2012 at 9:18pm.

B. How many minutes would a person have to be on facebook to be considered the top 5%? Round appropriately to the nearest minute. If you come out with a number like 100.2 minutes, you would have to round up to 101 because 100 would not quite be in the top 5%

- statistics -
**PsyDAG**, Sunday, November 18, 2012 at 1:02pmFor one person:

Z = (score-mean)/SD

For 35 people"

Z = (score-mean)/SEm

SEm = SD/√n

In both cases, find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores.

B. Use same table to find .05 in the smaller area. Use Z value in first equation to find the score.

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