Transcript standard normal distribution - Kian

Continuous Probability
Distributions
Week 6
GT00303
CONTINUOUS RANDOM VARIABLE A random variable
that takes on an infinite number of values within a given
range. It is usually the result of some type of measurement.
EXAMPLES
 The length of each song on the latest Tim McGraw
album.
 The weight of each student in this class.
 The temperature outside as you are reading this book.
 The amount of money earned by each of the more
than 750 players currently on Major League Baseball
team rosters.
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We cannot list the possible values because there is an
infinite number of them.
Since there is an infinite number of values, the probability of
each individual value is virtually ZERO. E.g. P(X=2) = 0,
P(X=5) = 0
Hence, in a continuous setting, point probabilities are
ZERO.
It is more meaningful to determine the probability of a range
of values. E.g. P(X ≤ 2) or P(2 ≤ X ≤ 5).
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Probability Density Function (PDF)
A function f(x) is called a probability density function
(over the range a ≤ X ≤ b if it meets the following
requirements:
•
f(x) ≥ 0 for all between a and b, and
f(x)
area=1
a
•
b
x
The total area under the curve between a and b is 1.0.
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Uniform Distribution
The uniform probability distribution is perhaps the simplest
distribution for a continuous random variable.
f(x)
area=1
a
b
x
For PDF, the total area under the curve (line) is 1. So,
how much the height should be?
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Mean of the Uniform Distribution
SD of the Uniform Distribution
a+b
m=
2
s=
(b - a)
2
12
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Illustration 1:
The amount of gasoline sold daily at a service station is
uniformly distributed with a minimum of 2,000 gallons and a
maximum of 5,000 gallons.
Find the probability that daily sales will fall between 2,500
and 3,000 gallons.
f(x)
P(2500 ≤ X ≤ 3000)
2,000
5,000
x
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æ 1 ö
P 2500 £ X £ 3000 = 3000 - 2500 ç
= 0.1667
÷
è 3000 ø
(
) (
)
There is about 16.67% chance that between 2500 and 3000
gallons of gas will be sold on a given day.
What is the probability that the service station will sell at
least 4000 gallons?
P(X ≥ 4000)
f(x)
æ 1 ö
P ( X ³ 4000 ) = (5000 - 4000 ) ç
è 3000 ÷ø
= 0.3333
2,000
5,000
x
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What is the probability that the service station will sell
exactly 2500 gallons?
P(X = 2500)
f(x)
2,000
5,000
x
æ 1 ö
P ( X = 2500 ) = ( 2500 - 2500 ) ç
=0
÷
è 3000 ø
Hence, in a continuous setting, point probabilities are
ZERO.
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Illustration 2:
Southwest Arizona State University provides bus service to
students while they are on campus. A bus arrives at the North
Main Street and College Drive stop every 30 minutes
between 6 A.M. and 11 P.M. during weekdays. Students
arrive at the bus stop at random times. The time that a
student waits is uniformly distributed from 0 to 30 minutes.
How long will a student “typically” have to wait for a bus?
a + b 0 + 30
m=
=
= 15
2
2
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What is the standard deviation of the waiting time?
s=
( b - a)
12
2
=
(30 - 0)
12
2
= 8.66
What is the probability a student will wait more than 25
minutes?
P(25 < Wait Time < 30)
= (height)(base)
1
= (5)
30
= 0.1667
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What is the probability a student will wait between 10 and 20
minutes?
P(10 < Wait Time < 20)
= (height)(base)
1
= (10)
30
= 0.3333
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Normal Distribution
The normal distribution is the most important of all
probability distributions.
The normal distribution
is bell shaped and
symmetrical about the
mean
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The PDF of a normal distribution is given by:
The normal distribution is fully defined by two parameters:
its standard deviation and mean
Unlike the range of the uniform distribution (a ≤ x ≤ b), normal
distributions range from minus infinity to plus infinity
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The normal distribution is described by two
parameters: μ and σ.
Increasing
the
mean shifts the
normal curve to
the right.
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The normal distribution is described by two
parameters: μ and σ.
Increasing
the
standard deviation
flattens the normal
curve.
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Standard Normal Distribution
0
1
1
A normal distribution whose mean is zero and standard
deviation is one is called the standard normal distribution.
Any normal distribution can be converted to a standard normal
distribution with simple algebra. This makes statistical
inferences much easier.
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We can convert any normal random variable to a standard
normal random variable (Z variable) as follows:
This shifts the mean
of X to zero
0
Some
advice:
always draw a
picture!
This changes the
shape of the curve
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Area under the Normal Curve
The total area under the normal
curve is 1.00; half of the the area
(0.5) under the normal curve is to the
right of μ, and the other half (0.5) to
the left of μ.
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Illustration 3:
Suppose that at another gas station the daily demand for regular gasoline
is normally distributed with a mean of 1,000 gallons and a standard
deviation of 100 gallons.
The station manager has just opened the station for business and notes
that there is exactly 1,100 gallons of regular gasoline in storage.
The next delivery is scheduled later today at the close of business. The
manager would like to know the probability that he will have enough
regular gasoline to satisfy today’s demands.
X = Daily demand for regular gasoline (in gallons)
X N    1000,  =100
P(X < 1100) = ??
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P(X < 1100)
Convert
X
into
standard
normal
random variable (Z).
-3
-2
+1
-1
+2
+3

 X   1100  1000 
P  X<1100  = P 
<

100 
 
= P  Z  1
Use Standard Normal
Probability Table!
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P(Z < 1) = 0.8413
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Standard Normal Table
The standard normal table provides P(Z < z), that is
the total area to the left of z.
For standard normal table, we can’t find the
probability that Z is equal to a specific value of z, say
P(Z = 1.2).
This is because normal random variable is
continuous and the probability that a continuous
random variable is equal to any single value is ZERO
(Point probabilities are ZERO).
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P(Z < -1.52) = 0.0643
P(Z > 1.8) = 1 – P(Z < 1.8)
= 1 – 0.9641
= 0.0359
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P(-1.3 < Z < 2.1)
= P(Z < 2.1) – P(Z < -1.3)
= 0.9821 – 0.0968
= 0.8853
Can you get:
P(-2.10 < Z < 0) = 0.4821
P(0 < Z < 2.5) = 0.4938
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Illustration 4:
Consider an investment whose return is normally
distributed with a mean of 10% and a standard deviation
of 5%.
a. Determine the probability of losing money.
b. Find the probability of losing money when the
standard deviation is equal to 10%.
X = Return for an investment (in %)
X N    10%,  =5%
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Determine the probability of losing money.
P(X < 0) = ??
 X   0  10 
P  X< 0  = P 
<
 = P  Z  2
5 
 
P(Z < -2) = 0.0228
Find the probability of losing money when the standard
deviation is equal to 10%.
P(X < 0) = ??
 X   0  10 
P  X< 0  = P 
<
 = P  Z  1
10 
 
P(Z < -2) = 0.1587
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Finding Value of z
Given an area (A) under the curve, what is the
corresponding value of z (zA) on the horizontal axis
that gives us this area?
This is particularly useful when performing hypothesis
testing!
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Illustration 5:
In hypothesis testing, an important step is to find the
critical value. Suppose that this is a one-tail test and the
level of significance is 10%.
What is the critical value of z?
P(Z < z) = 0.90
z = 1.28
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Illustration 6:
The lifetimes of the heating element in a Heatfast electric oven
are normally distributed, with a mean of 7.8 years and a
standard deviation of 2.0 years.
a. If the element is guaranteed for 2 years, what is the
probability that the ovens sold will need replacement in the
guarantee period because of element failure?
b. Heatfast is reconsidering the length of the guarantee period
on heating elements. Calculate the length of guarantee
period such that Heatfast would expect to replace a
maximum of 1% of sets due to element failure.
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X = Lifetime of a heating element (in years)
X N    7.8,  =2
What is the probability that the ovens sold will need replacement in
the guarantee period?
P(X < 2) = ??
 X   2  7.8 
P  X< 2  = P 
<
 = P  Z  2.9 
2 
 
P(Z < -2.9) = 0.0019 @ 0.19%
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Calculate the length of guarantee period such that Heatfast would
expect to replace a maximum of 1% of sets due to element failure.
P(Z < z) = 0.01
0.0019
z = -2.33
X=2
0.01
z = ???
X = ???
What is the corresponding
X value?
X
Z

X  7.8
2.33 
2
X  3.14
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Exponential Distribution
The exponential distribution has this probability density function:
Characteristics:
1. Positively skewed, similar to the
Poisson
distribution
(for
discrete variables).
2. Described by only one
parameter, which we identify as
λ, often referred to as the “rate”
of occurrence parameter.
3. As λ decreases, the shape of
the distribution becomes “less
skewed.”
The exponential distribution usually describes
inter-arrival situations such as:
• The service times in a system.
• The time between “hits” on a web site.
• The lifetime of an electrical component.
• The time until the next phone call arrives in a
customer service center
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Illustration 7:
The lifetime of an alkaline battery (measured in hours) is
exponentially distributed with λ = 0.05
Find the probability a battery will last between 10 & 15 hours.
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P 10<X< 15  = P  X< 15   P  X< 10 
0.05 15 
  1  e 0.0510  
= 1  e
 

= e 0.0510   e 0.0515 
= e 0.5  e 0.75
= 0.6065  0.4724
= 0.1341
P(10<X<15)
There is about
13.41%
chance a battery will only last
10 to 15 hours
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Other Continuous Distributions
Student t Distribution
The degrees of
freedom () defines
the
Student
t
distribution.
As the number of degrees of freedom increases, the t
distribution approaches the standard normal distribution
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Chi-Squared (2) Distribution
The degrees of freedom
()
defines
the
2
distribution.
The 2 distribution starts at zero
(is non-negative) and is not
symmetrical.
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F Distribution
The degrees of freedom
()
defines
the
F
distribution.
The F distribution starts at
zero (is non-negative) and
is not symmetrical.
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