MGMT350
Decision Science
Continuous Probability Distributions
Probability 3-of-3
CSUEB R. Choy ©2022
Rev C1
Overview
Random Variables
Discrete Random Variables
Continuous Random Variables
Continuous Probability
Distributions
+∞
Descriptive Parameters
Uniform Probability Distribution
Triangular Probability Distribution
Normal Probability Distribution
Exponential Probability Distribution
න 𝑓 𝑥 𝑑𝑥 = 1
−∞
0 ≤ 𝑓 𝑥𝑖 ≤ 1
Key Concept: Differences
• Discrete random variables – the probability mass function
f(x) provides the probability that the random variable
assumes a particular value.
• Continuous random variables – the counterpart of the
probability mass function is the probability density function.
Key Concept: Density Function
• Counterpart of the probability mass function is the probability
density function, also denoted by f(x)
• The probability density function does not directly provide
probabilities
• For continuous random variables, the probability of any
particular value of the random variable is zero
• We are computing the probability that the random variable
assumes any value in an interval
Expected Value and Variance
• Measure of central location or central tendency
=𝐸 𝑥
+∞
= ‫׬‬−∞ 𝑥 ∗ 𝑓 𝑥
𝑑𝑥
• Measure of variability or dispersion
 = 𝑉𝐴𝑅 𝑥 = 𝐸[ (𝑋 − ) ]
Uniform Distribution
Density Function
Every interval of along the length being equally likely
𝑓 𝑥 =
1
𝑏 −𝑎
0
𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑏
𝐸𝑣𝑒𝑟𝑦𝑤ℎ𝑒𝑟𝑒 𝑒𝑙𝑠𝑒
Uniform Distribution
Cumulative Probabilities, Expected Value, Variance
• Cumulative Probabilities
𝑥0 − 𝑎
𝑃 𝑥 ≤ 𝑥0 =
𝑏−𝑎
𝑓𝑜𝑟 𝑎 ≤ 𝑥0 ≤ 𝑏
• Expected value and variance
𝐸 𝑥
𝑎+𝑏
=
2
Var 𝑥
(𝑏−𝑎)2
=
12
Uniform Distribution
Example: Equally Likely Contractor Project Times
a = 3 days, b =7 days
𝑃 𝑥 ≤ 4 𝑑𝑎𝑦𝑠
1
1
= (4-3)
=
7−3 4
𝑃 𝑥 > 4 𝑑𝑎𝑦𝑠 = 1 - P 𝑥 ≤ 4 𝑑𝑎𝑦𝑠
E(x) = (7+3)/2 = 5
(7−3)2 4
VAR(x)=
=
12
3
3
=
4
𝑥0 = 4
Triangular Distribution
• Useful only when subjective probability estimates are available!
Triangular Distribution
Density Function
2(𝑥−𝑎)
(𝑏−𝑎)(𝑚−𝑎)
𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑚
2(𝑏−𝑥)
(𝑏−𝑎)(𝑏−𝑚)
𝑓𝑜𝑟 𝑚 ≤ 𝑥 ≤ 𝑏
𝑓 𝑥 =
Triangular Distribution
Density Function
2(𝑥−𝑎)
(𝑏−𝑎)(𝑚−𝑎)
𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑚
2(𝑏−𝑥)
(𝑏−𝑎)(𝑏−𝑚)
𝑓𝑜𝑟 𝑚 ≤ 𝑥 ≤ 𝑏
𝑓 𝑥 =
Triangular Distribution
Cumulative Probability Function - Derivation
P 𝑋 ≤ 𝑥𝒐 =
𝒙𝟎
𝒙−𝒂 𝟐
(𝒃 − 𝒂)(𝒎 − 𝒂)
𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑚
Triangular Distribution
Probability Function
𝑃 𝑥 ≤ 𝑥𝑜
(𝑥0 −𝑎)2
(𝑏−𝑎)(𝑚−𝑎)
𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑚
𝑥𝑜
=
1 −
𝑏−𝑥0 2
𝑏−𝑎 𝑏−𝑚
𝑓𝑜𝑟 𝑚 ≤ 𝑥 ≤ 𝑏
𝑥𝑜
Triangular Distribution
Probability Function
𝑃 𝑥𝑜 ≥ 𝑥
(𝑥0 −𝑎)2
1−
(𝑏−𝑎)(𝑚−𝑎)
𝑓𝑜𝑟 𝑎 ≤ 𝑥 ≤ 𝑚
𝑏−𝑥0 2
𝑏−𝑎 𝑏−𝑚
𝑓𝑜𝑟 𝑚 ≤ 𝑥 ≤ 𝑏
𝑥𝑜
=
𝑥𝑜
Triangular Distribution
Density Function
𝑃 𝑥1 ≤ 𝑥𝑜 ≤ 𝑥2
?
𝑥1
𝑥2
Triangular Distribution
Expected Value & Variance
• Expected Value
𝐸 𝑥
𝑎+𝑏+𝑚
=
3
• Variance
𝑎2 + 𝑏2 + 𝑚2 − 𝑎𝑏 − 𝑎𝑚 − 𝑏𝑚
Var 𝑥 =
18
Triangular Distribution
Example 1: Estimates of task times
Best case, a = 3 days,
Worst case, b =12 days
Most likely case, m = 6 days
𝑃 𝑥 ≤ 5 𝑑𝑎𝑦𝑠
(5−3)2
4
=
=
(6−3)(12−3) 27
3+6+12
E(x) = 3 = 7
VAR(x)=
32 +62 +122 −3∗12−3∗6−6∗12
18
𝑥0 = 5
= 3.5
Triangular Distribution
Example 2: Estimates of task times
Best case, a = 3 days,
Worst case, b =12 days
Most likely case, m = 6 days
𝑃 𝑥 ≤ 8 𝑑𝑎𝑦𝑠
(12−8)2
= 1(12−6)(12−3)
=1-
16
54
≈ 0.704
𝑥0 = 8
Triangular Distribution
Example 2: Estimates of task times
Best case, a = 3 days,
Worst case, b =12 days
Most likely case, m = 6 days
𝑃 𝑥 ≥ 8 𝑑𝑎𝑦𝑠
=
(12−8)2
=
(12−6)(12−3)
16
54
≈ 0.296
𝑥0 = 8
Triangular Distribution
Example 1 & 2
Best case, a = 3 days,
Worst case, b =12 days
Most likely case, m = 6 days
𝑎+𝑏+𝑚
𝐸 𝑥 =
3
= (3+6+12)/3
= 7 days
𝑎2 + 𝑏2 + 𝑚2 − 𝑎𝑏 − 𝑎𝑚 − 𝑏𝑚
Var 𝑥 =
18
2
2
2
3 +12 +6 −3∗12−3∗6−12∗6
=
18
= 3.5 days2
Normal Distribution
Density function
1
𝑓(𝑥)=
𝑒
 2𝜋
− 𝑥−𝜇
2
2
Z-Score
𝑥−µ
𝑧=
𝜎
=Norm.Dist(X,Mean,Standard_dev,Cumulative)
Cumulative: True for cumulative, False for density

2
Normal Distribution
Example: Hand Crafted Goods
Unit production: normally distributed
Mean: 8 hours, Std. Dev.: 1.5 hours
Published lead time: 10 hrs.
What is the probability of not being
late?
Excel:
=norm.dist(10,8,1.5,True)
=0.9088
𝑥0 = 10
Normal Distribution Using Tables
Example: Hand Crafted Goods
Z-Score
10−8
4
𝑥−µ
𝑧=
=
=
≈
1.33
𝜎
1.5
3
Z = 1.33 standard deviations above the mean.
Normal Distribution Example Problem 2
• Soda filling process…
• Assume a filling process for 2 liter bottles has a true population mean
of 2000 ml of liquid.
• The standard deviation of the process is 20 ml.
• If the fill liquid exceeds 2060 ml, it will overflow.
• What is the probability of an overflow?
• If the process involves 1,000,000 units /day (opportunities), then
what is the expected number of failures (overflows) each day?
• How can one improve the outcome?
Exponential Distribution
Density Function
Exponential Distribution
Density and Cumulative Function Equations
𝑓 𝑥
1 −𝑥/𝜇
= 𝑒
𝜇
𝑓𝑜𝑟 𝑥 ≥ 0
𝐹 𝑥 = 𝑃 𝑥 ≤ 𝑥0 = 1 − 𝑒 −𝑥0/𝜇
Exponential Distribution Example
Example: Pizza Delivery Service Guarantee (30 min. or it is FREE!)
If delivery times are exponentially distributed with a mean of 20
minutes...
What is the probability of not being late?
𝑃 𝑥 ≤ 30)
20 = 1 − 𝑒 −30/20 = 0.777
What is the probability of being late?
𝑃 𝑥 > 30)
20 = 𝑒 −30/20 = 0.223
Exponential Distribution Example
Example: Pizza Delivery Service Guarantee (30 min. or it is FREE!)
If delivery times are exponentially distributed with a mean of 20
minutes...
What is the probability of not being late?
𝑃 𝑥 ≤ 30)
20 = 1 − 𝑒 −30/20 = 0.777
What is the probability of being late?
𝑃 𝑥 > 30)
20 = 𝑒 −30/20 = 0.223
Exponential Distribution P&L
Example: Pizza Delivery Service Guarantee (30 min. or it is FREE!)
Based on the prior calculations, 77.% of deliveries would be on
time or early, while 22.3% would be late.
If a non-late delivery yields $8 in profit, while a late delivery
results in a $5 loss, then what is the long term expected unit
profit or loss?
Expected Profit (Loss) = 0.777*$8 + 0.223*(-$5) = +$5.10
Summary