A.

SPECIALTY TOYS CASE

Managerial Report

1. Use the sales forecaster’s prediction 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.
• From the case, we can know：
o the demand distribution can approximate a normal probability distribution.
o an expected demand of 20,000 units with a .95 probability that demand would be
between 10,000 units and 30,000 units.

Specialty Toys Demand Distribution

f(x)

0.95

0.025 0.025

z
0 5000 10000 15000 20000 25000 30000 35000 40000

Hence: The demand distribution is a normal random probability distribution with:

o normal random variable x.

o mean μ = 20000.
o standard deviation σ.
o P(10000 ≤ x ≤ 30000) = 0.95

• Calculating Standard deviation (σ):

P(x ≥30000) = P(x ≤ 10000) = 0.025

P(20000≤x≤30000) = 0.475.

Using the standard normal probability table, we see that the standard normal random
variable z = 1.96 when P (z = 1.96) = 0.475.

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 To calculate for 𝜎: at x= 30,000

𝑥−𝜇 30,000 − 20,000

𝜎= = = 5,102
𝑧 1.96

Hence, the standard deviation is 5,102.

• Standardized Normal Distribution (µ= 0, σ=1):

Specialty Toys Demand Distribution

f(x)

z
-3 -2 -1.96 -1 0 1 1.96 2 3

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2. Compute the probability of a stock-out for the order quantities suggested by members of the
management team.
a. If order quantity is 15,000:

15,000 − 20,000
𝑧= = −0.98
5102

f(x)
𝑃(𝑍 ≥ −0.98) = 0.3365 + 0.5 = 0.8365

0 z
-3 - -0.98
-1 0 1 1.9 3

b. If order quantity is 18,000:

18,000 − 20,000
𝑧= = −0.392
5102

f(x)
𝑃(𝑍 ≥ −0.392) = 0.1517 + 0.5 = 0.6517

-0.392 00 z
-3 - -1 1 1.9 3
c. If order quantity is 24,000:

24,000 − 20,000
𝑧= = 0.784
5102
f(x)

𝑃(𝑍 ≥ 0.784) = 0.5 – 0.2823 = 0.2177

0 0.784 z
d. If order quantity is 28,000: -3 - -1 0 1 1.9 3

28,000 − 20,000
𝑧= = 1.57
5102
f(x)

𝑃(𝑍 ≥ 1.57) = 0.5 – 0.44179 = 0.0582

0 1.57 z
-3 - -1 0 1 1.9 3

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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 Sales Cost ($16) Sales Amount Profit
Quantity Quantity
At $24 Surplus: At $5
15,000 10,000 $240,000 $25,000 $25,000
20,000 $360,000 0 $120,000
$240,000
30,000 $360,000 0 $120,000
18,000 10,000 $240,000 $40,000 -$8,000
20,000 $288,000 $432,000 0 $144,000
30,000 $432,000 0 $144,000
24,000 10,000 $240,000 $70,000 -$74,000
20,000 $384,000 $480,000 $20,000 $116,000
30,000 $576,000 0 $192,000
28,000 10,000 $240,000 $90,000 -$118,000
20,000 $448,000 $480,000 $40,000 $72,000
30,000 $672,000 0 $224,000

4. 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?
• Find x.

𝑥−𝜇 𝑥 − 20,000
𝑧= = = 0.52
𝜎 5102

x = 𝜎z + µ = 5102 (0.52) + 20,000 = 22,653

f(x)

70% 30%
0 0.52 z
-3 - -1 0 1 1.9 3
(20,000) (22,653)

Hence, the order quantity for meeting a demand at 70% and having stock-outs at 30% is 22,653.

• Projected profit at 22,653 order quantity:

Order Sales Cost ($16) Sales Amount Profit

Quantity Quantity
At $24 Surplus: At $5
22, 653 10,000 $240,000 $63,625 -$59,183
20,000 $480,000 $13,265 $130,817
$368, 488
30,000 $543,672 0 $181,224

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5. Provide your own recommendation for an order quantity and note the associated profit projections.
Provide a rationale for your recommendation.
• According to case and the above calculation, we get:
o underage cost (Cu) = 24-16 = 8.
o overage cost (Co) = 16-(16-5) = 5.
• According to the formula:
Cu
P(z)=
Cu + Co
• Critical ratio in normal distribution:

8
P(z)= = 0.62.
8+5
• In the standard normal distribution function table, we find:

P (z ≤ 0.31) = 0.5+0.12172= 0.62172

• 0.62172 is the probability value closest to 0.62.

• According to the formula:

x -μ
• z= .
σ
• Round up rule:

x= z *σ+μ= 0.31×5102.041+20000 = 21582

• Hence, we would recommend to Specialty Toys to have an order quantity of 21,582

because it will maximize the company’s profit.

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B. MARION’S CASE
1. Calculate the probability that the mean score of Blugert given by the simple random sample of
Marion Dairies customers will be 75 or less.
• Find the value of z.
Given: µ=80 σ=25 𝑥̅ = 75 n=50

𝑥̅ − 𝜇 75 − 80
𝑧= 𝜎 = = −1.41421
25
√𝑛 √50

• By using the standard normal table for the value 1.41, we get .42073

𝑷( 𝒛 ≤ −𝟏. 𝟒𝟏) = 0.5 – 0.42073 = 0.07927

f(x)

0.07927
0.07927 0.42073

z
-3 -
-1.41 -1 00 1 1.9 3

Hence, the probability that the mean score of 50 random customers is 75 or less is 7.93%.
Therefore, the possibility that Strawgurt becomes a potential product for those who don’t
like Blugurt is low.

2. If the Marketing Department increases the sample size to 150, what is the probability that the mean
score of Blugert given by the simple random sample of Marion Dairies customers will be 75 or less?
• If n= 150
𝑥̅ − 𝜇 75 − 80
𝑧= 𝜎 = = −2.45
25
√𝑛 √150

• By using the standard normal table for the value 2.45, we get .49286

𝑷( 𝒛 ≤ −𝟐. 𝟒𝟓) = 0.5 – 0.49286 = 0.00714

f(x)

0.00714 0.49286

z
-3 -
-2.45 -1 00 1 1.9 3
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 Hence, the probability that the mean score of 50 random customers is 75 or less is .71%.

3. Explain to Marion Dairies senior management why the probability that the mean score of Blugert
given by the simple random sample of Marion Dairies customers will be 75 or less differs for these
two sample sizes.
• The probabilities for these two simple random samples differ because of the sample size,
which affect the standard error. When the sample size is 30, the standard error is 3.53, but
when the sample size is 150 the standard error becomes lower, 2.04. It means that when the
sample size becomes larger, the standard error is lower so we have a better estimation of the
population mean. This principle is according to the Central Limit Theorem. For a sample size
150, the probability that the mean score of Blugert given by the simple random sample of
Marion Dairies customers will be 75 or less is very low, so we can suppose that the mean score
will be 75 or more, that confirms the consumers’ rating of the product with mean equals 80.