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
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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
Rounded to the closest hundreth, the standard deviation for the set is approximately 19.52. Juxtaposed
Standard Deviation for the mean column is 0.476Standard Deviation for the median column is 0.754Standard deviation for the mean column has least variability
In our second assumption, instead of using the cost of goods per cases in 1986, we try to use the percentage it counts in the total expenses which is 50.4% and to find the sales needed to break-even. The detail of the calculation is shown in the answer for questions d. The result is that 95,635, a little bit higher than the estimated sales of 90,000.
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Compute the projected profit for the order quantities suggested by the management team under three scenarios: worst case in which sales = 10,000 units, most likely case in which sales = 20,000 units, and best case in which sales = 30,000 units.
An intelligence test for which the scores are normally distributed has a mean of 100 and a standard deviation of 15. Use this information to describe how the scores are distributed.
Afterwards use the standard deviation equation to plug in your number's to get the standard deviation
Since the expected demand is 2000, thus, the mean µ is 2000. Through Excel, we get the z value given a 95% probability is 1.96. Thus, we have: z= (x-µ)/ σ=(30000-20000)/
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values under the curve lie within one standard deviation of the mean and 95% lie within two standard deviations3. This is good for intervallic continuous variables.
Standard deviation based on the available information (January to June) is 41.83 or approximately 42.