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

The average amount of time people spend on facebook each day is 71 minutes, with a standard deviation of 4.7 minutes. Are you more likely to select a random person that spends less than 68 minutes per day, or a group of 35 people that spend on average less than 65 minutes per day on facebook? Assume this is normally distributed. Find the chances of each to decide which is more likely, and by how much. Show all work.

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 1:02pm
For 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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