Let X have a gamma(?,?) with ? = r/2 , where r is a positive integer, and ? = 2. We say that X has a chi-square distribution with r degrees of freedom (df), denoted ?2(r). The family of chi-square distributions has special significance which becomes - PowerPoint PPT Presentation

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Let X have a gamma(?,?) with ? = r/2 , where r is a positive integer, and ? = 2. We say that X has a chi-square distribution with r degrees of freedom (df), denoted ?2(r). The family of chi-square distributions has special significance which becomes

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Section 3.5 Let X have a gamma( , ) with = r/2 , where r is a positive integer, and = 2. We say that X has a chi-square distribution with r degrees of freedom ... – PowerPoint PPT presentation

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Title: Let X have a gamma(?,?) with ? = r/2 , where r is a positive integer, and ? = 2. We say that X has a chi-square distribution with r degrees of freedom (df), denoted ?2(r). The family of chi-square distributions has special significance which becomes


1
Section 3.5
Let X have a gamma(?,?) with ? r/2 , where r is
a positive integer, and ? 2. We say that X has
a chi-square distribution with r degrees of
freedom (df), denoted ?2(r). The family of
chi-square distributions has special significance
which becomes apparent with the study of some
other distributions (such as the normal
distribution). (Table IV in Appendix B of the
textbook provides certain values of the
distribution function for selected ?2(r) random
variables and is designed especially for
applications to be discussed later.) The p.d.f.
of X is f(x) E(X) Var(X) The m.g.f. of
X is
xr/2 1 ex/2 if x gt 0 ?(r/2) 2r/2
r
2r
1 for t lt 1 / 2 (1 2t)r/2
M(t)
2
1. (a)
Flaws in a certain type of recording tape follow
a Poisson distribution and occur at an average
rate of 3 flaws per 5000 ft. The following
random variables are defined X number of
thousands of feet until the first flaw is
observed, Y number of thousands of feet until
the second flaw is observed, V number of
thousands of feet until the 6th flaw is observed.
Identify the type of distribution that each of X,
Y, and V has.
X has an distribution
exponential(5/3) (a gamma(1,5/3)
distribution).
Y has a distribution.
gamma(2,5/3)
V has a distribution.
gamma(6,5/3)
3
(b)
Find each of the following P(X gt 2)
?
?
3e 3x / 5 dx 5
e 3x / 5
e 6 / 5 0.301
2
x 2
or Define the random variable W number of flaws
in 2 thousand feet . W has a Poisson(6/5)
distribution.
P(X gt 2) P(W 0)
0.301
(from the Poisson Distribution Table)
4
P(Y lt 4)
Define the random variable W number of flaws in
4 thousand feet . W has a Poisson(12/5)
distribution.
4
9y e 3y / 5 dy or 25
0
P(Y lt 4) P(W ? 2) 1 P(W ? 1)
0.692
(from the Poisson Distribution Table)
P(V lt 9)
Define the random variable W number of flaws in
9 thousand feet . W has a Poisson(27/5)
distribution.
9
729v5 e 3v / 5 dv or 1875000
0
P(V lt 9) P(W ? 6) 1 P(W ? 5)
0.454
(from the Poisson Distribution Table)
5
2. (a)
Flaws in a certain type of wire follow a Poisson
distribution and occur at the average rate of 5
per thousand feet. Note that in order to use the
chi-square tables to obtain probabilities
concerning a gamma random variable, we must have
? 1/? 2, that is, ? 1/2. Since flaws occur
at an average rate of 5 per thousand feet, we can
also say that flaws occur at an average rate of
1/2 0.5 per hundred feet. From the chi-square
tables, what can be said about the probability
that
more than 1244 feet of wire will be used before
the 10th flaw,
Define the random variable X number of
hundreds of feet until the 10th flaw is
observed. X has a gamma(10,2) distribution, which
is a ?2( ) distribution.
20
P(X gt )
12.44
1 P(X ? 12.44)
0.90
6
(b) (c)
more than 3141 feet of wire will be used before
the 10th flaw,
Using the random variable X defined in part (a),
we have
P(X gt )
31.41
1 P(X ? 31.41)
0.05
less than 1831 feet of wire will be used before
the 5th flaw,
Define the random variable Y number of
hundreds of feet until the 5th flaw is
observed. Y has a gamma(5,2) distribution, which
is a ?2( ) distribution.
10
P(Y lt )
18.31
P(Y ? 18.31)
0.95
7
(d) (e)
between 777.9 and 948.8 feet of wire will be used
before the 2nd flaw,
Define the random variable W number of
hundreds of feet until the 2nd flaw is
observed. W has a gamma(2,2) distribution, which
is a ?2( ) distribution.
4
P( lt W lt )
7.779
9.488
P(W ? 9.488) P(W ? 7.779)
0.95 0.90 0.05
more than 1900 feet of wire will be used before
the 4th flaw,
Define the random variable V number of
hundreds of feet until the 4th flaw is
observed. V has a gamma(4,2) distribution, which
is a ?2( ) distribution.
8
P(V gt )
19
1 P(V ? 19)
between 0.01 and 0.025
8
(f)
less than 3600 feet of wire will be used before
the 9th flaw,
Define the random variable Q number of
hundreds of feet until the 9th flaw is
observed. Q has a gamma(9,2) distribution, which
is a ?2( ) distribution.
18
P(Q lt )
36
P(Q ? 36)
greater than 0.99
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