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Unformatted text preview: 113 Chapter3 Continuous Distributions SPECIAL
CONTINUOUS
DISTRIBU
TIONS 3.29 For each of the following functions: (i) ﬁnd the constant c so thalﬂx) is a p.d.f., {it}
ﬁnd the distribution function Fair), (iii) sketch the graphs of the p.d.f. f(x) and the
distribution function Fix). 3 (a) f(x) = (b) f(x) = Cﬁ. C
(0) f(x) = (d) N) = Let X have the p.d.f.f(x) = (x +1)/2. —1 < x < 1. Find
(a) No.64.
(b) 6h = No.25.
(C) 7mm The logistic distribution is associated with the distributicm function 17(1) : (1 + 5’)“. #00 c .t < 00. Find the p.d.l. of the logistic distribution and show.
that its graph is symmetric about x = 0. (I < I d: c.
0 : x : 4.
G : x 4: 1. Is this p.d.f. bounded? leIcoo. One hundred observations of X having a certain distribution were taken and then
ordered. yielding 0.19 0.25
0.48 0.53
0.83 0.85
1.02 1.03
1.28 1.30
1.49 1.50
1.61 1.62
1.71 1.71
1.81 1.84
1.91 1.92 0.26
0.54
0.87
1.05
1.34
1.51
1.62
1.72
1.85 1.85
1.92 1.93 (2) Calculate the sample mean Y. (b) Calculate the sampie standard deviation 5. (c) Group these data into 10 classes with class boundaries 0.005, 0.205, 0.405, ,_
2.005. (d) Construct a relative frequency histogram and then Buperimpose the p.d.f.
f(x) = #2. D < x < 2. Is there a good ﬁt? (e) Find ,u. and n. assuming that the p.d.f. of X isf(x} = 111.0 < x s 2. Is If close to
#715 5 close to a? 0.26
0.57
0.90
1.07
1.34
1.53
1.62
1.73 0.28
0.66
0.91
1.13
1.37
1.54
1.63
1.73
1.86
1.93 {1.31
0.69
0.94
1.13
1.39
1.56
1.64
1.74
1.87
1.93 0.39
0.70
0.96
1.14
1.40
1.56
1.66
1.75
1.87
1.94 0.42
0.72
0.99
1.20
1.43
1.57
1.67
1.76
1.88
1.95 0A4
0.73
1.01
1.22
1.43
1.60
1.70
1.77
1.90
1.98 0.45
0.79
1.02
1.27
1.46
1.60
1.71
1.80
1.91
1.99 Let the random variable X denote the outcome when a point is selected at random
from an interval [a,b], oo < a < b < 00. If the experiment is performed in a fair.
manner, it is reasonable to assume that the probability that the point is seieetetif
from the interval [a,x]. a 5 x < b is (x — a},r'(b — a). That is. the probability =
proportional to the length of the interval so that the distribution function of X is 0, x<a, x—a F(x)= , a5x<b, bea
1. bsx. Because X is a continuoustype random variable, F’(x) is equal to the p.d.f. of '
whenever F’(x) exists; thus when a < x < b, we have f(x) = F’(x) = 1/(b — a). EXAMPLE Section 3.3 Special Continuous Distributions 119 The random variable X has a uniform distribution if its p.d.f. is equal to a
constant on its support. In particular, if the support is the interval [a, b], then 1 f(x)= b_a, a 5 x 5 b.
Moreover, we shall say that X is U(a,b). This distribution is also referred to as
rectangular because the graph off(x) suggests that name. See Figure 3.3—1 for the
graph 0ff(x) and the distribution function F(x) when a : 0.30 and b = 1.55. Note
that we could have taken f(a) = 0 or f(b) = 0 without altering the probabilities
since this is a continuous~type distribution, and we shall do this in some cases.
The mean and variance of X are not difﬁcult to calculate (see Exercise 3.31).
They are Mza+b 022(b—a)2' 2 1.0 (a) Figure 3.31 Uniform p.d.f. and distribution function An important uniform distribution is that for which a = 0 and b = 1, namely
U(0, 1). If X is U(0, 1), approximate values of X can be simulated on most comput
ers using a random number generator. In fact, it should be called a pseudorandom
number generator because the programs that produce the random numbers are
usually such that if the starting number (the seed number) is known, all subsequent
numbers in the sequence may be determined by simple arithmetical operations.
Yet, despite their deterministic origin. these computerproduced numbers do
behave as if they were truly randomly generated, and we shall not encumber our
terminology by adding pseudo. Examples of computerproduced random numbers
are giVen in the Appendix in Table VIII. Treat each of these numbers as a decimal;
that is, divide each of them by 104. Let X have the p.d.f.
1 f(x) = 100, 0 < x < 100, so that X is U(0, 100). The mean and the variance are 2 _ (100 — 0)2 _ 10,000 :0+100=50 and a M 2 12 12 120 Chapter3 Continuous Distributions The standard deviation is a : 10mm, which is 100 times that of the U(0,1)
distribution. This agrees with our intuition since the standard deviation is a measure of spread and U(0, 100) is clearly spread out 100 times more than Uth, 1).
_— We say that the random variable X has an exponcnllal distribution if its p.d.f. i
of the form 1 r
f(vY)=geﬂ/9, 051<Ooa where the parameter (9 > 0. The mean of X is
so 1 _ 6 s _ a m
[1 : :V/O x x/ [1x : [_xe—I{ _ Be H ]o = 6 and E 2 00 .2 1 —.\‘/0 _ 2 ~.\‘/9 —x/6 1 —x,x‘6 w __ 2
(X )= ,t a 6 dx __ [—x e — {21)99 r — 29: e L _ 26 .
0 Thus the variance of X is
02 = E(X2) — n2 = 292 _ 91 = 92.
The distribution function of X is 0, ~oo<x<0, f(W)dW:[ 1—c’1/‘9, 05x<oo. See Example 5.27 for calculations of these characteristics using Maple.
The p.d.f. and distribution function are graphed in Figure 3.3—2 for 3 = 5. Thai
median, m, is found by solving Hm) = 0.5. That is, 1 — [We = 0.5. m : —61n(0.5) = 6 luff}. Fm
1 .(l 0.8
0.6
0.4 0.2 Figure 3.3—2 Exponential pill. and distribution function EXAMPLE EXAMPLE Section 3.3 Special Continuous Distributions 121 So with 9 = 5, the median is m = ~51n(0.5) = 51n(2) = 3.466. Both the median
and the mean 9 = 5 are indicated on the graphs.
It is useful to note that for an exponential random variable, X, we have that P(X>x> = 1—F(x)=1—(1_e—x/e) = e—x/b‘, when x > 0. Customers arrive in a certain shop at a mean rate of 20 per hour. What is the
probability that the shopkeeper will have to wait more than 5 minutes for the
arrival of the ﬁrst customer? Let X denote the waiting time in minutes until
the ﬁrst customer arrives and assume that X is exponentially distributed with the
mean 9 = 60/20 = 3. Thus f(x)=%€_(1/3)x, 0 E x < oo. 00
P(X > 5) = f g e(1/3>xdx = 633/3 = 0.1889.
5 The median time until the ﬁrst arrival is m = —31n(0.5) = 31n(2) = 2.0794. Suppose that the life of a certain type of electronic component has an exponential
distribution with a mean life of 500 hours. If X denotes the life of this component
(or the time to failure of this component), then 00 1 4/500 4/500
8 d! = e ' ' . P(X>x)=/; ﬂ Suppose that the component has been in operation for 300 hours. The conditional
probability that it will last for more than 600 additional hours is P({X > 900} n {X > 300})
W _ PlX > 900)
‘ P(X > 300)
ew900/500 = {400/500 : e P(X> 900lX> 300): —6/5 It is important to note that this conditional probability is exactly equal to
PU! } 600) = (5/5. That is, the probability that it will last longer than an
additional 600 hours, given that it has operated 300 hours, is the same as the
probability that it would last longer than 600 hours when ﬁrst put into operation.
Thus, for such components, an old component is as good as a new one, and we
say that the failure rate is constant. Certainly, with constant failure rate, there is 122 Chapter3 Continuous Distributions no advantage in replacing components that are operating satisfactorily. Obviously,
this is not true in practice because most would have an increasing failure rate with
time; hence the exponential distribution is probably not the best model for the probability distribution of such a iife.
__.— Before determining the characteristics of the gamma distribution, let us consider
the gamma function for which the distribution is named. The gamma function is:
deﬁned by i‘(t) 2/ y’”le“y dy, 0 < t.
o This integral is positive for 0 < 1' because the integrand is positive. Values of it are2
often given in a table of integrals. If: 3 1, integration of the gamma function of t
by parts yields F(t) = [—y1’”16?‘y:l;o + [0000 —— 1)y’_ze~y dy
= (t — 1) foo/y’ae‘v dy = (t —— 1)I‘(t — 1).
0 For example, 115) =. SETS) and [13} = 2F(2) = (2)(1)I‘(1). Whenever I = n, a
positive integer, we have, by repeated application of F(t) = (t — 1)f‘(t — 1), that F(n)=(tz—1)F(n —— 1) = (n — 1)(n — 2) .  ~ (2)(1)F(1). However, 00
r(1)=f e""dy = 1.
0 Thus, when n is a positive integer, we have that
I‘(n) = (n —— 1)! and, for this reason, the gamma function is called the generalized factorial.
Incidentally, F(1) corresponds to 0!, and we have noted that F(1) = 1, which is..
consistent with earlier discussions. The random variable X has a gamma distribution if its p.d.f. is of the form 1
N )9“ x0464”, 0 5 x < 00,
oz f(X) = where the parameters or and 0 are positive. (See Exercise 3.312.) Note that the!
exponential distribution is a special case of the gamma distribution when a = 1.
To determine the mean of the gamma distribution, integrate 1 F( )9“ xa—ie~x/6dx
(X n=E(X)=[0mx Section 3.3 Special Continuous Distributions 123 by changing variablesy 2: x16 or. equivalently, x 2 9y so dx/dy = 6. We obtain 0c 1 a_ _V
info (mummy) 1eedy ‘9 00 <+1>1
2—— a _ —ydl
' I‘M/o y e l _ 6F{a +1) _ 9aF(a) _
‘ r(a) ‘ ma) _ v using the property F(a + 1) = aF(a) of the gamma function. Likewise E(X2), by
the same change ofvariables, equals F(a + 2) __ 62(01 + 1)(a)l"(a) W) =92—rw — W) = 01(01 + 1)62. Thus the variance of X is
(72 = a(a +1)t92 ~ (m9)2 = 062. These characteristics can also be calculated using Maple. (See Example 5.2—8 .) Remark If the instructor El the second putt of Section ELI m: the
m.g.f. of ii mnﬂnttuus :andmn amiable can um: be studied. In order to see the effect of the parameters on the shape of the p.d.f., several
combinations ofo: and 6 have been used for graphs thatare displayed in Figure 3.3—3. f(x)
025 a: 1/4 0.20
0.15
0.10 0.05 (b)
Figure 3.3—3 Gamma p.d.f.s: 6 = 4,a =1/4,1,2,3,4;a = 4, 6 = 5/6, l.2,3,4 We now consider a special case of the gamma distribution that plays an important
role in statistics. Let X have a gamma distribution with 6 = 2 and a = r/2, where r
is a positive integer. The p.d.f. of X is 1 _ _.
f(X)= Wilfrﬂ 16 1/2, 05x <00. 124 Chapter3 Continuous Distributions EXAMPLE EXAMPLE We say that X has a chlsquare distribution m'th r degrees ol' freedom, WhiCh we
abbreviate by saying X is x2(r). The mean and the variance of [his chisquare
distribution are u=a6=(:>2=r and 2 02 = (162 :2 (:>22 2 2r. 2 That is. the mean equals the number of degrees of freedom, and the variance
equals twice the number of degrees of freedom. An explanation of “number nf
degrees of freedom” is given later. In Figure 3.34 the graphs of chisquare p.d.f.s for r = 2,3,5, 8, and 14 are glVCIL
Note the relationship between the mean. ,tt 2 r, and the point at which the p.d.f.
obtains its maximum. Bacause the chi—square distribution is so important in applications, tables have
been prepared giving the values of the distribution function X 1
F : r/2—l iw/Zd
(x) /0 ————F(r/2)2rx2 w e w for selected values of r and x. For an example, see Table IV in the Appendix. Figure 3.3—4 Chisquare p.d.f.s withr=2,3,5,8,14 2 4 6 8 1012141613 20 22 24 Let X have a chisquare distribution with r = 5 degrees of freedom. Then, using
Table IV in the Appendix, P(1.145 5 X 5 12.83): F(12.83) — 1:11.145): 0.975 — 0.050 = 0.925 P(X > 15.09) = 1 ~ F(15.09) = 1 — 0.99 = 0.01. IfX is X20), two constants, a and b, such that
P(a s: X < b) = 0.95 are a = gamma) = 1.690‘and b = 1030250) 2 16.01, where, in general, x30) is the _
point in a x20) distributiOn with a probability to the right of it. Other constants. a Section 3.3 Special Continuous Distributions 125 and b can be found. and we are only limited in our choices by the table. It is very
easy to ﬁnd these other constants using Minitab, Maple, or some other statistical package. . EXERCISES 3.3 3.31 3.32 3.33 Show that the mean and the variance of the uniform distribution Ma, b) are _ 2
=a+b and 022(1) a). 2 12
Let f(.l') = 1/2. —l 5 x 5 1, be the p.d.f. of X. Graph the p.d.f. and distribution
function. and record the mean and variance of X.
Let 1’ have a uniform distribution Unit). 1), and let ILL W:a+(b—a)Y, a<b. (3) Find the distribution function of W. HINT: Find P[a + (b — a)Y 5 w].
(b) How is W distributed? The Holland Sentinel reported the following numbers of calls per hour received by
911 between noon, February 26. and all day February 27. 303491622576422410
313342133232521024 They also reported the following lengths of time per minute between calls. 30 17 65 8 38 35 4 19 7 14 12 4 5 4 2
7 5 12 50 33 10 15 2 10 1 5 30 41 21 31
1 18 12 5 24 7 6 31 1 3 2 22 I 30 2
1 3 12 12 9 28 6 50 63 5 17 11 23 2 46 90 13 21 55 43 5 19 47 24 4 6 27 4 6 37 16 41 68 9 5 28 42 3 42 8 52 2 11 41 4 35 21 3 17 10 16 1 68 105 45 23 5 10 12 17 (a) Show graphically that the numbers of calls per hour have an approximate Poisson
distribution with El mean of it = 3. (b) Show that the sample mean and the sample standard deviation of the times
between calls are both approximately equal to 20. (c) If X is an exponential random variable with mean 9 = 20, compare P(X > 15)
with the proportion of times that are greater than 15. (d) Compare P(X > 45.5  X > 30.5) with the proportion ofobservations that satisfy
this condition. Let X equal the time (in minutes) between calls that are made over the public safety
radio. On four different days (February 14, 2]. 28. and March 6) and during a period
of one hour on each day, the following observations of X were made: 57820172248864
34210185784510 (3) Calculate the values of the sample mean and sample standard deviation. Are
the}.I close to each other in value? (b) Construct a boxandwhisker diagram. Does it indicate that the data are skewed,
as would be true for observations from an exponential distribution? Let X have an exponential distribution with mean 9.
(a) Find the ﬁrst quartile, ()1. (b) How far is the ﬁrst quartile below the mean?
(c) Find the third quartile, q}. (d) How far is the third quartile above the mean? 126 Chapter3 Continuous Distributions THE NORMAL
DISTRIBUTION 3.37 The waiting times in minutes until two calls to 911, as reported by the Holland
Sentineion November 13 between noon and midnight, were: 20 28 81 4 9 41 9 11 10 24 20
44 18 30 16 53 15 38 50 84 44 69 Could these times represent a random sample from a gamma distribution with or = 2
and I9 = 120/7? {a} Compare the distribution and sample means. (b) Compare the distribution and sample variances. (cl Compare P(X < 35) with the proportion of times that are less than 35. (d) If possible, make some graphical comparisons. to) What is your conclusion? IfX is 112(5), determine the constants c and d so that P(c < X < d) = 0.95 and
P(X : (r) = 0.025. That is. ﬁnd c = “3975(5) and d = X53256). Let X have a gamma distribution with parameters a = 2 and 6. Ifx = 2 is the mode
of the distribution, ﬁnd 6 and P(X < 9.488). Let X have the uniform distribution U(0,1). Find the distribution function and then
the p.d.f. of Y = —21nX. HINT: Find P(Y g y) = P(X :_ 6‘13”) whenD : y < 00. Find the uniform distribution U(d, b) that has the same mean and variance as a X2(8)
distribution. Letﬂir} be the p.d.f. of a gamma distribution. Show that (foﬂx) dx = 1 by changing
variablesy = x/Q. The normal distribution is perhaps the most important distribution in statistical
applications since many estimators, like ‘1’!” for p and X for it have approximate
normal distributions. One explanation of this fact is the role of the normal
distribution in the Central Limit Theorem. One form of this theorem will be
considered in Section 3.6. In addition, some measurements, but not all, have
approximate normal distributions. As a matter of fact, the authors do not like it
when instructors “grade on the curve” because usually test scores are not normally
distributed!
The random variable X has a normal distribution if its p.d.f. is deﬁned by 1 exp ( )2 00 Jr < 00
‘_.____ _ < ‘
bat/271 2b: where a and b are parameters satisfying —oo c n c: on, 0 < I; < co, and also where
exp[v] means (3“. In the formula for ﬁx) for this normal p.d.f.. the symbols a and b
are the mean and the standard deviation of X. In Example 5211] we use Maple to
show this and also that f(x) is indeed a p.d.f. Thus we usually write the normal p.d.f. as f (X) = 1 .x =
f( ) 2N
and say that the random variable X with this p.d.f. is N(n,a2). If Z is N(0,1).
we shall say that Z has a standard normal distribution. Moreover, the distribution
function of Z is
¢(Zl — P(Z < z} “ f2 ——1 e‘“’2”2dw —. ~00 «J 271' I 348 Appendix c Answers to {JuddNumbered Exercises 3.3.“: (a) /.L = 1/2,a"~’ :1/20;
(b) u = 2,02 does not exist. (c) neither the mean nor the variance exists. 0, x < 0,
(i) c = 2, (ii) F(x) = x4/16, 0 g x < 2,
1. 2 5x : m. D. x as 0.
(11311311, 0 5 x .c 4,
1 , 45x<oo. (i) c = 3/16, (ii) m) = l (i) c = 1/2, this p.d.f. is unbounded, (ii) F(x) = l 0, x<1,
1—1/x, 15x<oo. (i) c = 1, (ii) F(x) =[ f(x) = e""/(1+e_")2. —00 < x < 00; ex (e—x)2 _ e—x ' (cs02 ‘ zfm f(—x) = G(w) = (w — a)/(b — a), a 5 w 5 b;
U(a,b). 10.524; 9.320; yes. 22 = 240/7 2 34.286, .7 = 32.636; 02 = 28, 800/49 = 587.755, 52 = 548.338;
0.605 m 0.591 = 13/22. 6 = 2, 0.950. a=8—4«/3,b=8+4«/3. (2)0.3078; (b)0.4959; (00.2711; (d) 0.1646;
(e) 0.0526; (f) 0.3174(03173); (g) 0.0456 (0.0455); (h) 0.0026 (0.0027). 0.3849; (b) 0.5403; (c) 0.0603; (d) 0.0013;
0.6826; (f) 0.9544; (g) 0.9974; (h) 0.99. ...
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This note was uploaded on 12/28/2010 for the course ECON 41 taught by Professor Guggenberger during the Fall '07 term at UCLA.
 Fall '07
 Guggenberger

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