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```Continuous Probability
Distribution
(Lesson - 03/C)
Dealing with Uncertainty on a
Grand Scale
Continuous Distributions
• In discrete random variable distributions,
there is always a gap between the points of
a distribution.
For example: the observable numbers on a
dice roll can only be 0, 1, 2, 3, 4, 5, or 6 but
NOT 4.2 or 3.5, etc.
• In a continuous distribution, there are no
gaps between values
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Continuous Distributions
•
Discrete
0.3
Continuous
0.3
0.25
0.25
0.2
0.2
0.15
p(x) 0.15
f(x)
0.1
0.1
0.05
0.05
0
0
0
1
2
3
4
5
6
7
8
9
10
-0.05
X
0
1
2
3
4
5
6
7
8
9
10
X
The distribution of a CRV is characterised
by a probability density function f(x)
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Continuous Distributions
•
•
•
•
•
•
•
Uniform Distribution
Normal Distribution
z-Distribution
t-Distribution
F-Distribution
Chi-Square Distribution
etc.
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Continuous Distributions
• The probability density function f(x) must
satisfy two conditions;
1
f(x)  0
(i.e. non negative)
2
The total area under the curve is 1
• As the area of a line is zero, the specific
value of any x value is always 0
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Continuous Distributions (Cont.)
There are 3 basic types of parameters for
Continuous Distributions:
– Shape (some may have more than one
shape parameter)
– Scale (specifies the size along the
horizontal axis - contract/expend)
– Location (specifies the location relative
to zero on the horizontal
axis).
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Uniform Distribution
The sales of a company varies between
\$1,000 and 2,000 with equal probability.
What is the the probability that the sales
revenue will be less than or equal to
\$1,300?
To answer this question we need to know
how to put in use the Uniform Distribution.
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The Uniform Distribution
• Could be used to model:
– when little knowledge about a
random variable is available.
• Characteristics:
– between minimum (a) and maximum (b)
– outcomes are equally likely
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Uniform Distribution
• A uniform distribution occurs when the
values of a random variable are distributed
evenly across the domain of the variable
0.12
0.1
0.08
0.06
0.04
0.02
0
0
1
2
3
4
5
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7
8
9
9
Uniform Distribution (Cont.)
a=
b=
f(x) =
F(x) =
F(x) =
Minimum
x = Specified Value
Maximum
= 1 / (b-a)
if a ≤ x ≥ b
Cumulative probability observing at most ( x )
= (x-a) / (b-a) if a ≤ x ≥ b
=1
if x ≥ b
=0
if x ≤ a
Expected value = (a+b)/2 ; Var = [(b-a)^2] / 12
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Example: Uniform Distribution
The sales of a company varies uniformly
between \$1,000 and \$2,000. What is the
probability that the sales revenue will be less
than or equal to \$1,300?
Specified Value
Minimum Value
Maximum Value
x=
a=
b=
The probability that the sales revenue
will be \$1,300 or less is
F(x) 30%
Dr. C. Ertuna
1,300
1,000
2,000
11
Excel: Uniform Distribution
1/A
B
C
D
E
F
G
H
2 Specified Value
x = 1,300 Specified Revenue
3 Minimum Value
a = 1,000 Minimum Sales Revenue
4 Maximum Value
b = 2,000 Maximum Sales Revenue
5
6 Prob getting ≤ x
7
8
=(D2-D3)/(D4-D3)
F(x) = 30.00% probability that
sales will be x or
less
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The Normal Distribution
The customer’s demand is normally
distributed with a mean of 750 units/month
and a standard deviation of 100 units/month
What is the probability that the demand
will exceed 900 units/month?
To answer this question we need to know
how to put in use the Normal Distribution.
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The Normal Distribution
• The most important distribution that you
distribution
• There are several characteristics that make
the normal distribution very important for
statisticians
It is an excellent approximation to many
populations. (Adult human heights and weights).
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Normal Distribution
• Could be used to model:
– many natural phenomena,
– deviation from specifications (quality control).
• Characteristics:
– Symmetrical and bell-shaped
– Most observations are close to the mean with
– Can be determined entirely by the values of  and 
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Normal Distribution (Cont)
• A typical normal distribution;
It could completely
be described by
μ and σ
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Normal Distribution (Cont)
• Two normal distributions with same σ but
different μ
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Normal Distribution (Cont)
• Two normal distributions with same μ but
different σ
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Normal Distribution (Cont.)
x=
Specified Value
μ=
Mean
σ2 = Variance
f(x) =
e^[-(x-μ)2/2σ2] / (2π σ2)1/2
F(x) = Cannot be expressed mathematically
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Normal Distribution (Cont)
• For a normal distribution, the following
rules hold true;
• 68.3 % of all observations lie within 1
standard deviation of the mean
• 95.4 % of all observations lie within 2
standard deviations of the mean
• 99.7 % of all observations lie within 3
standard deviations of the mean
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Normal Distribution (Cont.)
68.3 % of all
observations lie within
1 standard deviation of
the mean
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Normal Distribution (Cont.)
NORMDIST(x, mean, std_dev, cumulative)
x=
=
=
Specified value
Standard deviation
mean
x
Std_dev
mean
Expected value = E(x) =  ; Var = 2
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Example: Normal Distribution
The customer’s demand is normally distributed
with a mean of 750 units/month and a standard
deviation of 100 units/month. What is the
probability that the demand will exceed 900?
Specified Value
x=
Mean
μ=
Standard Deviation (=square root of variance) σ =
The probability that the demand will be
more than 900 units is σ
Dr. C. Ertuna
900
750
100
F(x) 6.68%
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Excel: Normal Distribution
1/A
B
C
D
E
F
G
H
2 Specified Value
x = 900
Specified Demand
3 Mean
μ = 750
Mean Demand
4 Standard_dev
σ = 100
Standard Deviation of Demand
5
6
7 Prob getting > x
8
=NORMDIST(D2,D3,D4,TRUE)
= 0.933
F(x) = 6.68% probability that
demand will be
more than x
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The Standard Normal Distribution
• A special case of the normal distribution,
the standard normal distribution has a
mean of 0 and a standard deviation of 1
0.5
0.4
0.3
0.2
4
3
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
0
-3
0.1
-4
f(z)
• The corresponding
standard random
variable is denoted
by Z
0.6
Z
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Standard Normal Distribution (Cont)
• Any normal distribution can be converted to a
The Standard Normal Distribution, simply by
converting it’s mean to 0 and it’s standard
deviation to 1. ie Subtracting  from each
observation and dividing by .
Z
X 

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Plot
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Exponential Distribution
The mean time to fail of a light bulb is 12,000
hours.
What is the probability that the light bulb
will fail before 5,000 hours?
To answer this question we need to know how
to put in use the Exponential Distribution.
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Exponential Distribution
• Could be used to model:
– events that recur randomly over time,
– customer arrivals (to a service system).
• Characteristics:
– Bounded below zero
– Greatest density at 0
– Only scale parameter λ, no shape or location
parameters
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Exponential Distribution (Cont.)
EXPONDIST(x, lambda, cumulative)
Specified value
=
1/ mean
f(x) =
ex
-x
F(x) = 1- e
x=
x
lambda
x >= 0
x >= 0
E(x) = 1/ ; Var = (1/)2
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Example: Exponential Distribution
The mean time to fail of a light bulb is 12,000
hours. What is the probability that the light
bulb will fail before 5,000 hours?
Specified Value
Mean
Lambda = 1 / mean
x=
μ=
=
The probability that the light bulb will
fail before 5,000 hours is
F(x) 34.08%
Dr. C. Ertuna
5000
12000
1/12000
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Excel: Exponential Distribution
1/A
B
C
D
2 Specified Value
x = 5,000
3 Lambda
=
E
F
G
H
Specified Demand
0.00008 Mean Demand
4
5
6 Prob getting < x
7
8
=EXPONDIST(D2,D3,TRUE)
F(x) = 34.08%
probability that
the bulb will fail
before than x
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Summary
Probability distributions help managers in:
• Drawing inferences and conclusions about
sample data and
• Building useful decision models that
incorporate uncertainty and risk
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Next Lesson
(Lesson - 04/A)
Statistical Sampling &Analysis of
Sample Data
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