Question

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 2-score computation to 2 decimal places and your final answer to 2 decimal places.) Duration

Answer 1

Solution :

Given that ,

mean = $\mu$ = 4

standard deviation = $\sigma$ = 0.40

a)

P( 4.00<4.60) =P[( 4.00-4)/ 0.40 < (x - $\mu$ ) /$\sigma$  < ( 4.60 - 4 ) /0.40 ]

= P(0< z < 1.5)

= P(z <1.5) -P(z<0)

Using standard normal table

=0.9332 - 0.5 = 0.4332

Probability = 0.4332

b)

P(x >4.60) = 1-p(x< 4.60)

=1- p [(x - $\mu$ ) / $\sigma$ < (4.60-4 /0.40)

=1- P(z <1.5)

= 1 - 0.9332 = 0.0668

probability = 0.0668

c)

P( 4.60<5.50) =P[( 4.60-4)/ 0.40 < (x - $\mu$ ) /$\sigma$  < ( 5.50 - 4 ) /0.40 ]

= P(1.5 <Z <3.75)

= P(z <3.75)- P(z < 1.5)

Using standard normal table

=0.9999 - 0.9332 = 0.0667

Probability = 0.0667

d)

P( 3.50<5.50) =P[( 3.50-4)/ 0.40 < (x - $\mu$ ) /$\sigma$  < ( 5.50 - 4 ) /0.40 ]

= P(-1.25<z<3.75)

= P(z <3.75)- P(z <-1.25)

Using standard normal table

=0.9999 - 0.1056 = 0.8943

Probability = 0.8943

e)

P(Z > z ) = 0.04

1- P(z < z) =0.04

P(z < z) = 1-0.04 = 0.96

z = 1.751

Using z-score formula,

x = z * $\sigma$ + $\mu$

x = 1.75*0.40 + 4

x = 4.7

4.7 minutes