Chapter 8 - Continuous Probability Distributions

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Transcript Chapter 8 - Continuous Probability Distributions

Chapter 8
Continuous Probability
Distributions
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
8.1
Probability Density Functions…
Unlike a discrete random variable which we studied in
Chapter 7, a continuous random variable is one that can
assume an uncountable number of values.
 We cannot list the possible values because there is an
infinite number of them.
 Because there is an infinite number of values, the
probability of each individual value is virtually 0.
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8.2
Point Probabilities are Zero
Because there is an infinite number of values, the
probability of each individual value is virtually 0.
Thus, we can determine the probability of a range of values
only.
E.g. with a discrete random variable like tossing a die, it is
meaningful to talk about P(X=5), say.
In a continuous setting (e.g. with time as a random variable), the
probability the random variable of interest, say task length, takes
exactly 5 minutes is infinitesimally small, hence P(X=5) = 0.
It is meaningful to talk about P(X ≤ 5).
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
8.3
Probability Density Function…
A function f(x) is called a probability density function (over
the range a ≤ x ≤ b if it meets the following
requirements:
1) f(x) ≥ 0 for all x between a and b, and
f(x)
area=1
a
b
x
2) The total area under the curve between a and b is 1.0
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8.4
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
8.5
Uniform Distribution…
Consider the uniform probability distribution (sometimes
called the rectangular probability distribution).
It is described by the function:
f(x)
a
b
x
area = width x height = (b – a) x
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=1
8.6
Example 8.1(b)…
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.
f(x)
2,000
5,000
x
What is the probability that the service station will sell at
least 4,000 gallons?
Algebraically: what is P(X ≥ 4,000) ?
P(X ≥ 4,000) = (5,000 – 4,000) x (1/3000) = .3333
Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
8.7
The Normal Distribution…
The normal distribution is the most important of all
probability distributions. The probability density function of
a normal random variable is given by:
It looks like this:
Bell shaped,
Symmetrical around the mean
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…
8.8
The Normal Distribution…
Important things to note:
The normal distribution is fully defined by two parameters:
its standard deviation and mean
The normal distribution is bell shaped and
symmetrical about the mean
Unlike the range of the uniform distribution (a ≤ x ≤ b)
Normal distributions range from minus infinity to plus infinity
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8.9
Standard Normal Distribution…
A normal distribution whose mean is zero and standard
deviation is one is called the standard normal distribution.
0
1
1
As we shall see shortly, any normal distribution can be
converted to a standard normal distribution with simple
algebra. This makes calculations much easier.
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8.10
Normal Distribution…
The normal distribution is described by two parameters:
its mean and its standard deviation . Increasing the
mean shifts the curve to the right…
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8.11
Normal Distribution…
The normal distribution is described by two parameters:
its mean and its standard deviation . Increasing the
standard deviation “flattens” the curve…
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8.12
Calculating Normal Probabilities…
Example: The time required to build a computer is normally
distributed with a mean of 50 minutes and a standard
deviation of 10 minutes:
0
What is the probability that a computer is assembled in a
time between 45 and 60 minutes?
Algebraically speaking, what is P(45 < X < 60) ?
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8.13
Calculating Normal Probabilities…
P(45 < X < 60) ?
…mean of 50 minutes and a
standard deviation of 10 minutes…
0
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8.14
Calculating Normal Probabilities…
P(–.5 < Z < 1) looks like this:
The probability is the area
under the curve…
We will add up the
two sections:
P(–.5 < Z < 0) and
P(0 < Z < 1)
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0
–.5 … 1
8.15
Calculating Normal Probabilities…
How to use Table 3… [other forms of normal tables exist]
This table gives probabilities P(0 < Z < z)
First column = integer + first decimal
Top row = second decimal place
P(0 < Z < 0.5)
P(0 < Z < 1)
P(–.5 < Z < 1) = .1915 + .3414 = .5328
The probability time is between
45 and 60 minutes = .5328
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8.16
Using the Normal Table (Table 3)…
What is P(Z > 1.6) ?
P(0 < Z < 1.6) = .4452
z
0
1.6
P(Z > 1.6) = .5 – P(0 < Z < 1.6)
= .5 – .4452
= .0548
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8.17
Using the Normal Table (Table 3)…
What is P(Z < -2.23) ?
P(0 < Z < 2.23)
P(Z < -2.23)
P(Z > 2.23)
z
-2.23
0
2.23
P(Z < -2.23) = P(Z > 2.23)
= .5 – P(0 < Z < 2.23)
= .0129
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8.18
Using the Normal Table (Table 3)…
What is P(Z < 1.52) ?
P(0 < Z < 1.52)
P(Z < 0) = .5
z
0
1.52
P(Z < 1.52) = .5 + P(0 < Z < 1.52)
= .5 + .4357
= .9357
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8.19
Using the Normal Table (Table 3)…
What is P(0.9 < Z < 1.9) ?
P(0 < Z < 0.9)
P(0.9 < Z < 1.9)
z
0
0.9
1.9
P(0.9 < Z < 1.9) = P(0 < Z < 1.9) – P(0 < Z < 0.9)
=.4713 – .3159
= .1554
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8.20
Example 8.2
The return on investment is normally distributed with a
mean of 10% and a standard deviation of 5%. What is the
probability of losing money?
We want to determine P(X < 0). Thus,
 X   0  10 
P(X  0)  P


5 
 
 P( Z   2)
 .5  P(0  Z  2)
 .5  .4772
 .0228
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8.21
Finding Values of ZA…
Often we’re asked to find some value of Z for a given
probability, i.e. given an area (A) under the curve, what is
the corresponding value of z (zA) on the horizontal axis that
gives us this area? That is:
P(Z > zA) = A
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8.22
Finding Values of Z…
What value of z corresponds to an area under the curve of
2.5%? That is, what is z.025 ?
Area = .50
Area = .025
Area = .50–.025 = .4750
If you do a “reverse look-up” on Table 3 for .4750,
you will get the corresponding zA = 1.96
Since P(z > 1.96) = .025, we say: z.025 = 1.96
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8.23
Finding Values of Z…
Other Z values are
Z.05 = 1.645
Z.01 = 2.33
Will show you shortly how to use the “t-tables” with infinite
degrees of freedom to find a bunch of these standard values
for Zα
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8.24
Using the values of Z
Because z.025 = 1.96 and - z.025= -1.96, it follows that we can
state
P(-1.96 < Z < 1.96) = .95
The old Empirical Rule stated about 95% within + 2σ
P(-2 < Z < 2) = .95
From now on we will use the 1.96 number for this statement
unless you are just talking in general terms about how much
of a population in with + 2σ
Similarly
P(-1.645 < Z < 1.645) = .90
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8.25
Exponential Distribution…[Not on test]
Another important continuous distribution is the exponential
distribution which has this probability density function:
Note that x ≥ 0. Time (for example) is a non-negative quantity; the
exponential distribution is often used for time related phenomena such
as the length of time between phone calls or between parts arriving at
an assembly station. Note also that the mean and standard deviation are
equal to each other and to the inverse of the parameter of the
distribution (lambda )
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8.26
Exponential Distribution…[Not on test]
The exponential distribution depends upon the value of
Smaller values of
“flatten” the curve:
(E.g. exponential
distributions for
= .5, 1, 2)
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8.27
Other Continuous Distributions…
Three other important continuous distributions which will be
used extensively in later sections are introduced here:
Student t Distribution, Looks like the standard normal
distribution (Z) after someone sat on it
Chi-Squared Distribution,
F Distribution.
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8.28
Student t Distribution…[don’t really need to know formula]
Here the letter t is used to represent the random variable,
hence the name. The density function for the Student t
distribution is as follows…
(nu) is called the degrees of freedom, and
(Gamma function) is (k)=(k-1)(k-2)…(2)(1)
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8.29
Student t Distribution…[1 parameter]
In much the same way that
and
define the normal
distribution [2 parameters], , the degrees of freedom,
defines the Student [will use df]
t Distribution:
Figure 8.24
As the number of degrees of freedom increases, the t
distribution approaches the standard normal distribution.
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8.30
Using the t table (Table 4) for values…
For example, if we want the value of t with 10 degrees of
freedom such that the area under the Student t curve is .05:
Area under the curve value (t) : COLUMN
t.05,10
t.05,10=1.812
Degrees of Freedom : ROW
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8.31
Student t Probabilities and Values
Excel can calculate Student distribution probabilities and
values. Warning: Excel will give you the value for “t” where
 is the area in “BOTH” tails
=TINV(0.1,10) "="
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1.812
8.32
Chi-Squared Distribution…[Not on test]
The chi-squared density function is given by:
As before, the parameter
freedom.
is the number of degrees of
Figure 8.27
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8.33
F Distribution…[Not on test]
The F density function is given by:
F > 0. Two parameters define this distribution, and like
we’ve already seen these are again degrees of freedom.
is the “numerator” degrees of freedom and
is the “denominator” degrees of freedom.
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8.34
Problems: Standard Normal “Z”
If the random variable Z has a standard normal distribution,
calculate the following probabilities.
P(Z > 1.7) =
P(Z < 1.7) =
P(Z > -1.7) =
P(Z < -1.7) =
P(-1.7 < Z < 1.7)
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8.35
Problems: Normal Distribution
If the random variable X has a normal distribution with
mean 40 and std. dev. 5, calculate the following
probabilities.
P(X > 43) =
P(X < 38) =
P(X = 40) =
P(X > 23) =
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8.36
Problem: Normal
The time (Y) it takes your professor to drive home each
night is normally distributed with mean 15 minutes and
standard deviation 2 minutes. Find the following
probabilities. Draw a picture of the normal distribution and
show (shade) the area that represents the probability you are
calculating.
P(Y > 25) =
P( 11 < Y < 19) =
P (Y < 18) =
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8.37
Problem: Target the Mean
The manufacturing process used to make “heart pills” is
known to have a standard deviation of 0.1 mg. of active ingredient.
Doctors tell us that a patient who takes a pill with over 6 mg. of
active ingredient may experience kidney problems. Since you want to
protect against this (and most likely lawyers), you are asked to
determine the “target” for the mean amount of active ingredient in each
pill such that the probability of a pill containing over 6 mg. is 0.0035 (
0.35% ). You may assume that the amount of active ingredient in a pill
is normally distributed.
*Solve for the target value for the mean.
*Draw a picture of the normal distribution you came up with and show the 3
sigma limits.
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8.38