A study of longdistance phone calls made from General Electr

A study of long-distance phone calls made from General Electric\'s 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 3.8 minutes and the standard deviation was 0.70 minutes. (a) What fraction of the calls last between 3.8 and 4.5 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 4.5 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 4.5 and 5.5 minutes? (Round z-score computation to 2 decimal places and final answer to 4 decimal places.) Fraction of calls (d) What fraction of the calls last between 3.5 and 5.5 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% of the calls. What is this time? (Round z-score computation to 2 decimal places and your final answer to 2 decimal places.) Duration

Solution

A study of long-distance phone calls made from General Electric\'s corporate headquarters in Fairfield, Connecticut, revealed the length of the calls, in minutes,. follows the normal probability distribution.

mean length of time per call = 3.8

Standard deviation = 0.70

Round z-score computation to 2 decimal places and your final answer to 2 decimal places.

a)What fraction of the calls last between 3.8 and 4.5 minutes? using formula z = (x - mu )/s.d

P(3.8 < x < 4.5) = (4.5 - 3.8)/ 0.70 < x < (3.8 - 3.8)/0.70

= P(1 < z < 0 )

= 0.8413 - .5000

= 0.3413

b) (b) What fraction of the calls last more than 4.5 minutes?

p( x > 4.5) = P ( 1 - (x<4.5)

= P ( 1 - ( (4.5 - 3.8)/.7 )

= P ( 1 - (z = 1))

= p ( 1 - 0. 8413)

= 0.1587

c) What fraction of the calls last between 4.5 and 5.5 minutes?

p ( 4.5 < x < 5.5) = P ( (4.5 - 3.8) / 0.7 < x < ( (5.5 - 3.8) / 0.7))

= P( 1 < z < 2.43 )

= 0.9925 - 0.8413

= 0.1512

d)What fraction of the calls last between 3.5 and 5.5 minutes?

p ( 3.5 < x < 5.5) = P ( (3.5 - 3.8) / 0.7 < x < ( (5.5 - 3.8) / 0.7))

= P( - 0.43 < z < 2.43 )

= 0.9925 - 0.3336

= 0.6589

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% of the calls. What is this time?

Probablity of longest call =0.06

So probablity would be lie for longest call would be = 1 - 0.06 = 0.94

Calculating the z score using the z table = 1.56

Time of longest call = 1.56 = ( x - 3.8 ) / 0.70 = 4.89

   longest (in duration) 6% of the calls, time is = 4.89