Question 1081259: 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 4.20 minutes and the standard deviation was 0.40 minutes.
a.
What fraction of the calls last between 4.20 and 4.90 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 4.90 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 4.90 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) 6 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 for all of these
for 4.2 and 4.9
z=0/0.4=0 to (4.9-4.2)/0.4=0.7/0.4=1.75
probability from z between 0 and +1.75 is 0.4599.
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More than 4.9 minutes: We know that more than 4.2 is 0.5 and between 4.2 and 4.9 is 0.4599. Subtract the second from the first and get the value of greater than 4.9, which 0.0401. It is z> 1.75
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Between 4.9 and 5.5 minutes is z between +1.75 and 1.3/0.4=3.25, The probability is 0.0395.
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Between 3.5 and 5.5 minutes is z between -0.7/0.4=-1.75 and +3.25. That is 0.9594.
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The longest 6% of calls is where the probability is 0.9400. This is a z of +1.55.
From the top equation, z*sd=x-mean=1.55*0.4=.62=x-mean; x=4.2+0.62=4.82 minutes or 4m49s
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