Decision Support Tools: Exploring Expected Value, Sales Analysis, and Probability Calculations

In this document we will discuss about Decision Support Tools and below are the summary points of this document:- Expected value: Calculation of the expected value using the formula for discrete probability distribution. Table of sales: Detailed table showcasing sales units, expected value, variance, standard deviation, and more. Probability calculations: Determining probabilities for various scenarios using normal probability distribution and given data.

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Decision Support Tools
Student’s Name
Institution Affiliation

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Question One
a. Expected value
According to (Gravetter & Wallnau, 2016), the expected value is the mean value of a
random variable. It’s used to measure the average value of data.For the case of a discrete
probability distribution, the expected value of random variable X is computed by the
formula:
E ( X ) =∑ x∗px , where x=random variable , px=probability of x
The following is an example to illustrate this. The table below shows the distribution of the
number of children in a family. Required to compute the expected number of children in a
family
Number of Children(x) 1 2 3
Probability ( px) 0.6 0.3 0.1
E ( x ) =1 ( 0.6 ) +2 ( 0.3 ) +3 ( 0.1 )
¿ 0.6+ 0.6+0.3=1.5
≈ 2 children

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b. Table of sales of loaves
1. Completion of the table: Excel Output
Sales
Units(x)
Number
of days p(x)
Expected
Value
More
than
Less
than (x-E(x))^2
(x-E(x))^2
*p(x)
0 5 0.05 0.00 0.95 8.41 0.421
1 10 0.10 0.10 0.85 0.05 3.61 0.361
2 25 0.25 0.50 0.60 0.15 0.81 0.203
3 25 0.25 0.75 0.35 0.40 0.01 0.002
4 20 0.20 0.80 0.15 0.65 1.21 0.242
5 15 0.15 0.75 0.00 0.85 4.41 0.662
Total 100 1.00
E(x) 2.90 Variance 1.89
Standard
deviation 1.37
Excel Formula
Sales
Units(x)
Number of
days p(x)
Expected
Value
More
than
Less
than (x-E(x))^2
(x-E(x))^2
*p(x)
0 5 =B2/$B$9 =A2*C2 =1-C2
=(A2-
$D$10)^2 =G2*C2
1 10 =B3/$B$9 =A3*C3 =E2-C3 =1-E2
=(A3-
$D$10)^2 =G3*C3
2 25 =B4/$B$9 =A4*C4 =E3-C4 =1-E3
=(A4-
$D$10)^2 =G4*C4
3 25 =B5/$B$9 =A5*C5 =E4-C5 =1-E4
=(A5-
$D$10)^2 =G5*C5
4 20 =B6/$B$9 =A6*C6 =E5-C6 =1-E5
=(A6-
$D$10)^2 =G6*C6
5 15 =B7/$B$9 =A7*C7 =E6-C7 =1-E6
=(A7-
$D$10)^2 =G7*C7
Total
=SUM(B2:B
7)
=SUM(C2:
C7)
E(x)
=SUM(D2:D
7) Variance
=SUM(H2:H
7)
Standard
deviation
=SQRT(H10
)

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2. The average daily sales
2.90
3. The probability of selling 2 or more
0.85
4. The probability of selling 3 or less
0.65
5. The variance of the distribution
1.89
6. Standard deviation
1.37
c. The table of 10, 000 parts produced on four machines
Machine: W X Y Z All
Machine
s
Grade:
Satisfactor
y
3200 800 2400 1600 8000
Rework 600 150 450 300 1500
Scrap 200 50 150 100 500
All Grades 4000 1000 3000 2000 10000
The probability that a part selected at random:
1. Was produced by Machine and should be reworked.

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P= 600
10000 =0.06
2. Was produced by Machine Z and is not satisfactory.
P=1− 1600
10000 =1−0.16
¿ 0.84
3. Was produced by Machine Y and should be scrapped
P= 150
10000 =0.015
4. Needs Reworked
P= 1500
10000 =0.15
5. Needs to be scrapped given that it was produced by machine W
P= 200
4000 =0.05
d. The average sales of apples are 4000 with a standard deviation of 500
In this case, the probability of will be determined using the normal probability distribution
1. The Probability that sales will be greater than 4250 apples.
P=1−P (z < 4250−4000
500 )
¿ 1−P ( z <0.5 )=1−0.6915

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¿ 0.3085
2. The probability that sales will be less than 3600 apples
P=P ( z< 3600−4000
500 )=P ( z ←0.8 )
¿ 0.2119
3. The probability that sales will be less than 4500 apples.
P=P ( z< 4500−4000
500 )=P ( z <1 )
¿ 0.8413
Question Two: Research Question
1. What is the average age of the Australia population?
Is 37 years
2. What is the average age to die in Australia (for both men and women)?
The expected age to dies is 82.5 years.
3. What percentage of people work in Australia?
Is 62.4%
Question 3: Statistical Decision Making and Quality Control Show all calculations/reasoning
a. Company setting control limits for monitoring the direct labor
1. 95% confidence level

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μ= x ± z∗s
√ n
¿ 20 ± 1.96∗5
√64
¿ 20 ± 1.96∗5
8
¿ 20 ±1.225
Therefore, the 95% confidence is (18.775, 21.225), where UCL is 21.225 and LCL is 18.775
2. Control limit with a smaller sample of 9 observations
μ= x ± z∗s
√ n
¿ 20 ± 1.96∗5
√9
¿ 20 ±3.2667
Therefore, UCL=23.2667 and LCL is 16.7333
b. Hypothesis Testing
μ=42 years, σ =10.8 years ,α =0.05 , x=45
1. Hypotheses
H0 : μ=42
H1 : μ> 42
2. Z-statistics will be used, as the sigma (σ ¿ is known and the sample size (49) is
greater than 30

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z= x−μ
σ = 45−42
10.8 = 3
10.8
¿ 0.2778
The critical value of Z at the 0.05 significance level is 1.645. The z-statistic (0.2778) is
less than z-critical (1.645)
3. Sketch of the situation
4. Conclusion
The z-statistic (0.2778) is less than z-critical (1.645), therefore null hypothesis will be
accepted. The implication is that there’s no sufficient evidence to support the claim
of age discrimination as the test prove that the laying off was done at the average
age of worker (42 years).
Reference

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Gravetter, F. J., & Wallnau, L. B. (2016). Statistics for the behavioral sciences. Cengage
Learning.