Transcript Document

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.
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.
8.2
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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8.3
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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…
8.4
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
Normal distributions range from minus infinity to plus infinity
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8.5
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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8.6
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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8.7
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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8.8
Calculating Normal Probabilities…
We can use the following function to convert any normal
random variable to a standard normal random variable…
0
Some advice:
always draw a
picture!
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8.9
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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8.10
Calculating Normal Probabilities…
P(45 < X < 60) ?
…mean of 50 minutes and a
standard deviation of 10 minutes…
0
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8.11
Calculating Normal Probabilities…
OK, we’ve converted P(45 < X < 60) for a normal
distribution with mean = 50 and standard deviation = 10
to
P(–.5 < Z < 1) [i.e. the standard normal distribution with
mean = 0 and standard deviation = 1]
so
Where do we go from here?!
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8.12
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)
0
–.5 … 1
We can use Table 3 in Appendix B to look-up probabilities P(Z < z)
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8.13
Calculating Normal Probabilities…
Recap: The time required to build a computer is normally
distributed with a mean of 50 minutes and a standard
deviation of 10 minutes
What is the probability that a computer is assembled in a
time between 45 and 60 minutes?
P(45 < X < 60) = P(–.5 < Z < 1) = .5328
“Just over half the time, 53% or so, a computer will have
an assembly time between 45 minutes and 1 hour”
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8.14
Using the Normal Table (Table 3)…
What is P(Z > 1.6) ?
P(Z < 1.6) = .9452
z
0
1.6
P(Z > 1.6) = 1.0 – P(Z < 1.6)
= 1.0 – .9452
= .0548
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8.15
Using the Normal Table (Table 3)…
What is P(0.9 < Z < 1.9) ?
P(Z < 0.9)
P(Z < 1.9)
z
0
0.9
1.9
P(0.9 < Z < 1.9) = P(Z < 1.9) – P(Z < 0.9)
=.9713 – .8159
= .1554
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8.16
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 .9750,
you will get the corresponding zA = 1.96
Since P(z > 1.96) = .025, we say: z.025 = 1.96
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8.17
Finding Values of Z…
Other Z values are
Z.05 = 1.645
Z.01 = 2.33
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8.18
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
Recall that the empirical rule stated that approximately 95%
would be within + 2 standard deviations. From now on we
use 1.96 instead of 2.
Similarly
P(-1.645 < Z < 1.645) = .90
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8.19
Other Continuous Distributions…
Three other important continuous distributions which will be
used extensively in later sections are introduced here:
Student t Distribution, [will use in this class]
Chi-Squared Distribution,
F Distribution, [will use in this class]
Exponential.
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8.20
Student t Distribution…
Much like the standard normal distribution, the Student t
distribution is “mound” shaped and symmetrical about its
mean of zero: Looks like the normal distribution after an
elephant sat on it [flattened out/spread out more than a
normal]
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8.21
Student t Distribution…
In much the same way that
and
define the normal
distribution, , the degrees of freedom, defines the Student
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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8.22
Determining Student t Values…
The student t distribution is used extensively in statistical
inference. Table 4 in Appendix B lists values of
That is, values of a Student t random variable with
of freedom such that:
degrees
The values for A are pre-determined
“critical” values, typically in the
10%, 5%, 2.5%, 1% and 1/2% range.
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8.23
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 tail area under the Student t curve is
.05:
Area under the curve value (tA) : COLUMN
t.05,10
t.05,10=1.812
Degrees of Freedom : ROW
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8.24
F Distribution…
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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8.25
Determining Values of F…
For example, what is the value of F for 5% of the area under
the right hand “tail” of the curve, with a numerator degree of
freedom of 3 and a denominator degree of freedom of 7?
Solution: use the F look-up (Table 6)
There are different tables
for different values of A.
Make sure you start with
the correct table!!
F.05,3,7
F.05,3,7=4.35
Denominator Degrees of Freedom : ROW
Numerator Degrees of Freedom : COLUMN
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8.26
Problem: Normal Distribution
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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8.27
Problem: Normal
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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8.28
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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8.29
Problem: Normal – Targeting 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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8.30