654 Words3 Pages

1. Use the sales forecaster’s predication to describe a normal probability distribution that can be used to approximate the demand distribution. Sketch the distribution and show its mean and standard deviation.

Let's assume that the expected sales distribution is normally distributed, with a mean of 20,000, and 95% falling within 10,000 and 20,000.

We know that +/- 1.96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by:

z = (x - mu)/sigma = 1.96 sigma = (x - mu)/z

Sigma = (30,000-20,000) / 1.96 = 5,102 units.

So, we have a distribution with a mean of 20,000 and a standard deviation of 5,102.

2. Compute the probability of a stock-out for the order quantities*…show more content…*

One of specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. What quantity would be ordered under this policy, and what is the projected profit under the three sales scenarios?

The area under the curve to the left of the unknown quantity must be 0.7 (70%). So, we must first find the z value that cuts off an area of 0.7 in the left tail of standard normal distribution. Using the cumulative probability table, we see that z=0.53.

To find the quantity corresponding z=0.53, we have

Z=x-u/sigma. Were x is the required quantity.

x = 20,000 + 0.53(5102) = 22,704

The projected profits under the 3 scenarios are computed below for Order Quantity: 22,704.

|Unit Sales |Profit |

|10,000 |-58,968 |

|20,000

Let's assume that the expected sales distribution is normally distributed, with a mean of 20,000, and 95% falling within 10,000 and 20,000.

We know that +/- 1.96 standard deviations from the mean will contain 95% of the values. So, we can get the standard deviation by:

z = (x - mu)/sigma = 1.96 sigma = (x - mu)/z

Sigma = (30,000-20,000) / 1.96 = 5,102 units.

So, we have a distribution with a mean of 20,000 and a standard deviation of 5,102.

2. Compute the probability of a stock-out for the order quantities

One of specialty’s managers felt that the profit potential was so great that the order quantity should have a 70% chance of meeting demand and only a 30% chance of any stock-outs. What quantity would be ordered under this policy, and what is the projected profit under the three sales scenarios?

The area under the curve to the left of the unknown quantity must be 0.7 (70%). So, we must first find the z value that cuts off an area of 0.7 in the left tail of standard normal distribution. Using the cumulative probability table, we see that z=0.53.

To find the quantity corresponding z=0.53, we have

Z=x-u/sigma. Were x is the required quantity.

x = 20,000 + 0.53(5102) = 22,704

The projected profits under the 3 scenarios are computed below for Order Quantity: 22,704.

|Unit Sales |Profit |

|10,000 |-58,968 |

|20,000

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