Download Continuous Probability Distributions

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LECTURE 05
LECTURE 05 Continuous Probability Distributions
Engineering College, Hail University, Saudi Arabia
1 - Basic Concepts
1.1 Continuous Random Variables
A random variable X is continuous if its set of
possible values is an entire interval of
numbers (If A < B, then any number x
between A and B is possible).
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1.2 Continuous Probability Distribution
Lett X be
L
b a continuous
ti
rv. Then
Th
a probability
b bilit
distribution or probability density function (pdf)
of X is a function f (x) such that for any two
numbers a and b,
P ( a ≤ X ≤ b ) = ∫ f (x )dx
d
b
a
The graph of f is the density curve.
1.3 Probability Density Function
For f (x) to be a pdf
1.
f (x) > 0 for all values of x.
2. The area of the region between the graph of f and the
x – axis is equal to 1.
y = f ( x)
Area = 1
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Probability Density Function
P (a ≤ X ≤ b )
is g
given by
y the area of the
shaded region.
y = f ( x)
a
b
1.4 The Cumulative Distribution Function
The cumulative distribution function (cdf), F(x)
for a continuous rv X is defined for every number
x by:
F (x ) = P ( X ≤ x ) = ∫ f ( y )dy
x
−∞
For each x, F(x) is the area under the density
curve to the left of x.
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Using F(x) to Compute Probabilities
Let X be a continuous rv with pdf f(x) and cdf
F(x). Then for any number a,
P ( X > a ) = 1 − F (a )
and for any numbers a and b with a < b,
b
P ( a ≤ X ≤ b ) = F (b ) − F (a )
2 Continuous Distributions
Most experiments in operations have sample spaces
th t d
that
do nott contain
t i a fi
finite,
it countable
t bl number
b
off
simple events.
A distribution is said to be continuous when it is built
on continuous random variables, which are variables
that can assume the infinitely many values
corresponding to points on a line interval.
An example of a random variable would be the time it
takes a production line to produce one item.
In contrast to discrete variables, which have values
that are countable, the continuous variables’ values
are measurements.
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2 Continuous Distributions Types
The major continuous distributions
widely used in process operations
and quality engineering are:
• the exponential,
• the normal,
• the log-normal, and
• the Weibull distributions.
2.1- Exponential Distribution
The exponential distribution closely resembles the Poisson distribution.
The Poisson distribution is built on discrete random variables and
describes random occurrences over some intervals, whereas the
exponential distribution is continuous and describes the time between
random occurrences.
Examples of an exponential distribution are the time between machine
breakdowns and the waiting time in a line at a supermarket.
The exponential distribution is determined by the following formula:
The mean and the standard deviation are:
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2.1 - Exponential Distribution
The shape of the exponential
distribution is determined by only one
parameter, λ.
Each value of λ determines a different
shape of the curve.
The area under the curve between any
two points determines the probabilities
for the exponential distribution.
The formula used to calculate that
probability is:
2.1 - Exponential Distribution
Engineering Example (5.1)
Suppose that the time in months between line
stoppages
t
on a production
d ti
liline ffollows
ll
an exponential
ti l
distribution with λ = 5
a. What is the probability that the time until the line
stops again will be more than 15 months?
b. What is the probability that the time until the line
stops again will be less than 20 months?
c. What is the probability that the time until the line
stops again will be between 10 and 15 months?
d. Find μ and σ. Find the probability that the time until
the line stops will be between (μ − 3σ) and (μ + 3σ).
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2.1 - Exponential Distribution
Solution 5.1
Solution:
a)
e = 2.7182
The probability that the time until the line stops again will be more than
15 months is 0.000553.
b)
The probability that the time until the line stops again will be less than
20 months is 0.9999.
2.1 - Exponential Distribution
Solution 5.1
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2.1 - Exponential Distribution
Solution 5.1
2.2 - Normal Distribution
The normal distribution is one of the most widely used probability
distributions.
Most of nature and human characteristics are normally
distributed, and so are most production outputs for well calibrated
machines.
The normal probability is given by:
f (x) =
Where
2
2
1
e−( x−μ ) /(2σ ) −∞ < x < ∞
σ 2π
e = 2.7182828
π = 3.1416
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Normal Distribution
The equation of the distribution
depends on μ and σ.
The curve associated with that
function is bell-shaped and has an
apex at the center.
It is symmetrical about the mean,
and the two tails of the curve
extend indefinitely without ever
touching the horizontal axis
axis.
The area between the curve and
the horizontal line is estimated to
be equal to one.
Normal Distribution
Minitab Practice
Using Minitab functionalities
generate and plot normal
distributions with μ=0 and σ=1,
0.8, 0.6, 0.4, 0.3 and 0.1.
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Normal Distribution
Effect of μ and σ
1
2
3
Normal distributions with the
same mean but different
standard deviations.
deviations
4 Normal distribution with
different mean and different
standard deviation.
In Quality Control and
Improvement,
it
is
always recommended
to work on the process
to get:
1) μ= Target of the
Process
2) σ: smallest value
possible.
Normal Distribution
Effect of process mean μ and variability measure σ
Two normal curves having the same
standard deviation but different means.
Two normal curves having
different means and different
standard deviations.
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Two normal curves with the same mean
but different standard deviations.
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Normal Distribution Characteristics
The area under the curve is determined as shown on the graph.
68.26% of the area lies between μ − σ and μ + σ.
95.45% of the area lies between μ − 2σ and μ + 2σ.
99.73% of the area lies between μ − 3σ and μ + 3σ.
This characteristic has been used by quality professionals for process
capability analysis and process improvement.
Normal Distribution
Remember that the total area under
the curve is equal to 1, and half of
q
to 0.5.
that area is equal
• The area on the left side of any point
(A1) represents the probability of an
event being “less than” that point of
estimate (Z1),
• The area on the right (A2)
represents the probability of an event
being “more than” the point of
estimate (Z2).
(Z2)
• The shaded area (A3) under the
curve between a and b represents
the probability that a random variable
assumes a certain value in that
interval.
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Normal Distribution
Area Under Normal Distribution
Equation
Area
Z1
A1 = P ( Z ≤ Z 1) = Φ ( Z1) =
∫ f (Z ).dz
A1
−∞
A2 = P ( Z > Z 2) = 1 − Φ ( Z 2)
A3 = P ( Z 1 < Z ≤ Z 2) = Φ ( Z 2) − Φ ( Z 1)
Normal Distribution
Z-transformation:
The shape of the normal distribution depends on
two factors, the mean and the standard deviation.
Every combination of μ and σ represent a unique
shape of a normal distribution.
Based on the mean and the standard deviation, the
complexity involved in the normal distribution can
b simplified
be
i lifi d and
d it can b
be converted
t d iinto
t th
the
simpler z-distribution.
This process leads to the standardized normal
distribution,
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A2
A3
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Normal Distribution
Z table
Normal Distribution
Example. The weekly profits of a large group of stores are normally distributed
with a mean of μ = 1200 and a standard deviation of σ = 200.
a) What is the Z value for a profit for x = 1300? For x = 1400?
b) what is the percentage of the stores that make $1500 or more a week?
Solution:
a)
b)
From the Z score table, z=1.5 corresponds to 0.4332.
This represents the area between $1200 and $1500.
The area beyond $1500 is: 0.5 – 0.4332 = 0.0668
This area is 0.0668; i.e 6.68% of the stores make more than $1500 week.
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Normal Distribution
Z table Practical Applications
Find the area under the standard
normal curve for the following,
using the z-table. Sketch each
one.
(a) between z = 0 and z = 0.75
(b) between z = -0.55 and z = 0
(c) between z = -0.45 and z = 0.75
(d) between z = 0.45 and z = 1.50
(e) to the right of z = -1.35.
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Normal Distribution
Example 2 - Engineering Applications
It was found
f
d that
th t the
th mean length
l
th off 100 parts
t
produced by a lathe was 20.05 mm with a
standard deviation of 0.02 mm. Find the
probability that a part selected at random
would have a length
(a) between 20.03 mm and 20.08 mm
(b) between 20.06 mm and 20.07 mm
((c)) less than 20.01 mm
(d) greater than 20.09 mm.
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Normal Distribution
Example 2 – Solution
Normal Distribution
Example 2 – Solution
d) 20.09 is 2 s.d. above the mean, so the answer will be the same as (c),
P(X > 20.09) = 0.0228.
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Normal Distribution
Example 3 - Engineering Applications
The average life of a certain type of motor
i 10 years, with
is
ith a standard
t d dd
deviation
i ti off 2
years. If the manufacturer is willing to
replace only 3% of the motors that fail, how
long a guarantee should he offer? Assume
that the lives of the motors follow a normal
distribution.
Solution
X = life of motor
x = guarantee period
We need to find the value (in years) that will give us the bottom 3% of the
distribution.
These are the motors that we are willing to replace under the guarantee.
P(X < x) = 0.03
Normal Distribution
Example 3 - Engineering Applications
Solution
The area that we can find from the z-table is : 0.5 - 0.03 = 0.47
The corresponding z-score is z = -1.88.
Since
, we can write:
Solving this gives x = 6.24.
So the guarantee period should be 6.24 years.
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Normal Distribution
Laboratory Practice - 1
Normal Distribution
Laboratory Practice - Solution
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Normal Distribution
Laboratory Practice - Solution
Normal Distribution
Laboratory Practice - 2
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Normal Distribution
Minitab Capabilities.
From the Calc menu, select the “Probability distributions” option and then
select “Normal.”
Fill in
i the
h fields
fi ld in
i the
h “Normal
“N
l Distribution”
Di ib i ” dialog
di l box
b and
d then
h select
l
“OK.”
“OK ”
Th k You
Thank
Y
Any Questions ?
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Dr Mohamed AICHOUNI & Dr Mustapha BOUKENDAKDJI
http://faculty.uoh.edu.sa/m.aichouni/stat319/
Email: [email protected]
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