Specialty Toys Case 12121245
1. Let X 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
P(10000 < X < 20000) = 0.475
P( X < 10000)=0.025
NORM.S.INV(0.025)=-1.96
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
The graph above shows the distribution for the demand for the Weather Teddy Bear using Specialty Toys’ forecasts based off of sales histories for similar products. This forecast predicts that this toy will have a demand of 20,000 units. However, the forecasts also predict that the probability of selling between 10,000 to 30,000 units is equal to 0.95. Using this information, the forecast suggests a mean sale of 20,000 units with a range of 10,000 to 30,000 units. Using the normal standard inverse function in excel, the standard deviation for this forecast is calculated to be 5,102 units. Using 20,000 units as the given mean, this additional information can be used to generate the graph showing 14,898 and 25,102 units within one standard deviation of the mean, and 9,796 and 30,204 units within 2 standard deviations of the mean. 9,796 and 30,204 are outside of the 0.95 probabilty range.
2.
Order (X) z-score=(X-20000)/5102 P(X)
Stock out=P(1-X)
15000
-0.980
0.164
0.836
18000
-0.392
0.348
0.652
24000
0.784
0.783
0.217
28000
1.568
0.942
0.058
The above table shows the probability of stocking out for different scenarios based off of the four order quantities suggested by management. To derive these values, a z-score was determined for each scenario using the formula z= (X- μ)/ σ. For Orders of 15000, 18000, 24000, and 28000 with a mean of 20000 units, the z-scores are determined to be -0.98, -0.392, 0.784, and 1.568 respectively. Using the normal standard distribution fucntion in excel, the probality of these z-scores are calculated to be 0.164, 0.348, 0.783, and 0.942 respectively.
P(10000 < X < 30000) = 0.95
P( X < 20000)=0.5
P(10000 < X < 20000) = 0.475
P( X < 10000)=0.025
NORM.S.INV(0.025)=-1.96
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
The graph above shows the distribution for the demand for the Weather Teddy Bear using Specialty Toys’ forecasts based off of sales histories for similar products. This forecast predicts that this toy will have a demand of 20,000 units. However, the forecasts also predict that the probability of selling between 10,000 to 30,000 units is equal to 0.95. Using this information, the forecast suggests a mean sale of 20,000 units with a range of 10,000 to 30,000 units. Using the normal standard inverse function in excel, the standard deviation for this forecast is calculated to be 5,102 units. Using 20,000 units as the given mean, this additional information can be used to generate the graph showing 14,898 and 25,102 units within one standard deviation of the mean, and 9,796 and 30,204 units within 2 standard deviations of the mean. 9,796 and 30,204 are outside of the 0.95 probabilty range.
2.
Order (X) z-score=(X-20000)/5102 P(X)
Stock out=P(1-X)
15000
-0.980
0.164
0.836
18000
-0.392
0.348
0.652
24000
0.784
0.783
0.217
28000
1.568
0.942
0.058
The above table shows the probability of stocking out for different scenarios based off of the four order quantities suggested by management. To derive these values, a z-score was determined for each scenario using the formula z= (X- μ)/ σ. For Orders of 15000, 18000, 24000, and 28000 with a mean of 20000 units, the z-scores are determined to be -0.98, -0.392, 0.784, and 1.568 respectively. Using the normal standard distribution fucntion in excel, the probality of these z-scores are calculated to be 0.164, 0.348, 0.783, and 0.942 respectively.