Question

The time it takes a carrier to move goods from point A to point B follows a normal distribution with an average of 5.75 hours and a standard deviation of 1.2 hours.
5.1 Calculate the probability that a randomly selected job will take between 4.75 and 5.75 hours to be moved from point A to B. (4)
5.2 Calculate the probability that a randomly selected job will take less than 3.5 hours to be moved from point A to B. (5)
5.3 Calculate the probability that a randomly selected job will take longer than 5 hours to be moved from point A to B. (6)
5.4 Determine the minimum time it took to move the 15% of jobs that took the longest to be moved from point A to B.

Answer 1

we have given u = 5.75 and sigma =1.2

5.1 Calculate the probability that a randomly selected job will take between 4.75 and 5.75 hours to be moved from point A to B.

Answer : P [4.75 < x < 5.75 ] we know that if (x- u)/sigma then it follows standard normal distribution

= P [ ((4.75 - u) / sigma) < (x-u)/sigma < ((5.75 - u) /sigma ) ]

= P [ (4.75 - 5.75) /1.2 < z < (5.75 -5.75) / 1.2 ]

= P [ -0.8333 < z < 0 ]

= P [ z < 0 ] - P [z > -0.833]

= now use z table

= 0.5 - 0.2033

=0.2967 probability is 0.2967

5.2 Calculate the probability that a randomly selected job will take less than 3.5 hours to be moved from point A to B. (5)

= P [ x < 3.5 ]

= P [ z < (3.5 -5.75)/1.2]

= P [ z < -1.87]

= 1 - P [ z > -1.87 ]

=1 - 0.0307

= 0.9693 is probability .

5.3 Calculate the probability that a randomly selected job will take longer than 5 hours to be moved from point A to B. (6)

= P [ x > 5 ]

= P [ z > (5 - 5.75) / 1.2 ]

= P [ z > -0.625 ]

= 0.2676

5.4 Determine the minimum time it took to move the 15% of jobs that took the longest to be moved from point A to B.

Answer : here we have to find out x value

Z value for 15 % is 1.04 from z table

( x - u ) /sigma = 1.04

x = u + ( 1.04 * sigma)

x = 5.75 + ( 1.04 * 1.2)

x = 6.998