Question

Skinner Produce buys fresh Boston lettuce daily. Daily demand is normally distributed with a mean of 100 units and standard deviation of 15 units. At the beginning of the day Skinner orders 140 units of lettuce. What is the probability that Skinner will have at least 20 units left over by the end of the day?

a. 0.996

b. 0.91

c. 1.00

d. 0.09

Answer 1

Quantity ordered = 140 units

Quantity left at the end of the day = 20 units

Therefore, amount consumed that day = 140 – 20 = 120

Let Z value of the required probability that at least 20 units be leftover by end of the day = Z1

Therefore ,

Mean demand + Z1 x standard deviation of demand = 120

Or, 100 + Z1 x 15 = 120

Or, 15.Z1 = 20

Or, Z1 = 20/15 = 1.33

Corresponding probability for Z = 1.33 as derived from Z table is 0.90824 ( 0.91 rounded to 2 decimal places )

ANSWER : 0.91