Question

300 deliveries

1. If you are unable to change the standard deviation of the delivery time, what should be the average delivery time so that no more than 5% of the deliveries are late (late = takes more than 9 hours)

Current average delivery time = 8 hours, standard deviation of delivery time = 1 hour

Target average delivery time: _________ hours (round your answer to 2 decimal points)

2. If you are unable to change the average delivery time, what should be the standard deviation of the delivery process so that no more than 5% of the deliveries are late (late = takes more than 9 hours)

Current average delivery time = 8 hours, standard deviation of delivery time = 1 hour

Target standard deviation of delivery time: _________ hours (round your answer to 2 decimal points)

Answer 1

Answer to first question :

No more than 5% of the deliveries are late and target delivery time being 9 hours means probability that delivery time being 9 hours is 0.95 .

Corresponding z value for probability 0.95 = NORMSINV ( 0.95 ) = 1.6448

Let,

Target average delivery time = m

Standard deviation = sd = 1 day

Therefore,

M + Z x sd = 9

Or, m + 1.6448 x 1 = 9

Or, m + 1.6448 = 9

Or, m = 7.36 hours ( rounded to 2 decimal places )

Target average delivery time = 7.36 hours

Answer to 2^nd question :

Let ,

Target standard deviation of the delivery process = sd1

Mean delivery time = m = 8 hours

Probability that delivery time will not exceed 9 hours = 0.95

Corresponding z value for probability being 0.95 = NORMSINV ( 0.95 ) = 1.6448

M + Z x Sd1 = 9

Or, 8 + 1.6448xSd1 = 9

Or, 1.6448.Sd1 = 1

Or, Sd1 = 1/1.6448

Or, Sd1 = 0.6079 hours ( 0.61 hours )

Target standard deviation of delivery time = 0.61 hours