SOLUTION: A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal p

Question 1061671: A study of long-distance phone calls made from General Electric Corporate Headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes, follows the normal probability distribution. The mean length of time per call was 4.00 minutes and the standard deviation was 0.40 minutes.

a.
What fraction of the calls last between 4.00 and 4.60 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Fraction of calls

b.
What fraction of the calls last more than 4.60 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Fraction of calls

c.
What fraction of the calls last between 4.60 and 5.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Fraction of calls

d.
What fraction of the calls last between 3.50 and 5.50 minutes? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

Fraction of calls

e.
As part of her report to the president, the director of communications would like to report the length of the longest (in duration) 4 percent of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.)

Duration

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
z=(x-mean)/sd
Between 4 and 4.60 min is a z between 0 and 1.50, probability is 0.4332.
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Last more than 4.60 min is the difference between 0.5 and 0.4332. That is 0.0668.
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Between 4.60 and 5.50 minutes is z between +1.50 and +3.75 or 0.0667,
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Between 3.50 and 5.50 minutes is z between -1.25 and +3.75 or 0.8943.
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The 96th percentile is where z=+1.75
Multiply that by 0.40, the sd, and that is 0.70. Add that to the mean and the time is 4.70 min or 4m42s
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