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 suggested by members of the management team.

Using the normal distribution theory, we discover that as the ordered quantity increases the probability of stockout decreases.

At 15,000 the probability of stockout will be 0.8365
At 18,000 the probability of stockout will be 0.6517
At 24,000 the probability of stockout will be 0.2177
At 28,000 the probability of stockout will be 0.0582

3. 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:

Order Quantity: 15,000 were cost price is $16, selling price $24 & after holiday selling price $5 |Unit Sales |Profit |
|10,000 |25,000 |
|20,000 |120,000 |
|30,000 |120,000 |

Order Quantity: 18,000 were cost price is $16, selling price $24 & after holiday selling price $5 |Unit Sales |Profit |
|10,000 |-8,000 |
|20,000 |144,000 |
|30,000...

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...be the demand for the toy. Then X follows normal distribution with mean μ = 20000 and standard deviation σ. Then
P(10000 < X < 30000) = 0.95
P( X < 20000)=0.5
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NORM.S.INV(0.975)=1.96
Z-score of 10000 =-1.96
Z-score of 30000=1.96
σ = (30000-20000)/1.96 =10000/1.96 = 5102
Standard Deviation of 5102
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