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Unformatted text preview: 142 Chapter3 Continuous Distributions THE CENTRAL
LIMIT
THEOREM 3.57 The sample mean f is the maximum likelihood estimator 01'3 it the underlying
p.d.f. is f(x; 6) 2: (time—“f”, {J c .I' c on, where 19 > D. For this distribution,
E( Y ) = 6 and Van: Y ) := 92in, where n is the sample size. USe the fact that
T is approximately NW. 91,91] to construct an approximate 95 percent conﬁdence
interval for 9. Let .2? be the mean of a random sample of size If from a distribution with unknown
mean ,u and known variance 02. Then E :l: lafﬁ can serve as an approximate 95
percent conﬁdence interval for u. If a = 4, ﬁnd :1 so that this conﬁdence interval
is X :l: 0.5. In Section 2.5 we found that the mean Y of a random sample of size n from a
discrete distribution with mean it and variance 02 > 0 is a random variable with
the properties that 2
Var(3(_) = —.
n E(X)=a and Thus, as :1 increases, the variance of 7 decreases. Consequently, the distribution
of it" clearly depends on n, and we see that we are dealing with sequences of
distributions. Also note that the probability becomes concentrated in a small
interval centered at it. That is, as ii increases. I tends to converge to u, or
( it" ~ p.) tends to canverge to i] in a probability sense. When sampling from
a normal distribution, we will Show, in Chapter 6, that the distribution of f
is N(u,02/n).
In general, if we let it _Y—u Y—nu WW: ﬁa’ where Y is the sum of a random sample of size n from some distribution with mean
u and variance 02, then for each positive integer n, W: (Y—u) u—u _O/ﬁ_ E(W) = E[ o Y—u]=E(Y)—u 0N5 0N3 — ——= —1.
02/17 _ 2 _ (Y—MV _ EKY—mzl 02/" _
var(W) — E<W )5 E[—_] 02/11 02/11 Thus, while Y — u “degenerates” to zero, the factor ﬁ/a in JR 3? — u)/a
“spreads out” the probability enough to prevent this degeneration. What then is
the distribution of W as it increases? One observation that might shed some light
on the answer to this question can be made immediately. If the sample arises from
a normal distribution, we show in. Example 6.2—4 that ? is N(u,az/n) and hence
W is N(0,1) for each positive n. Thus, in the limit, the distribution of W must
be N (0, 1). So if the solution of the question does not depend on the underlying
distribution (i.e., it is unique), the answer must be N(U,1). As we will see, this
is exactly the case, and this result is so important it is called the Central Limit
Theorem, the proof of which is given in Section 6.3 . Section 3.6 The Central Limit Theorem 1143 Theorem 361 [Emmi Lima Theorem: It? is the: mean of a random sample x1, x1, . . ..x,,
al‘ sin n from a distribution with a ﬁnite mean 11 and a ﬁnite punitive variance
U2. then the distribution of
II
E In — n.“
r“! f—u_ _ Y~npi :1: w’ﬁ F ﬁn «ﬂier '
when; l" 3 EL] 1:}, is mu. 1] in Ihelimit 35:: —~ out W: A practical use of the Central Limit Theorem is approximating, when n is
“sufﬁciently large,” the distribution function of W, namely 67%;"2 dz = <D(w). W 1
P(W 5 w) w/ 00 271' We present some illustrations of this application, discuss “sufﬁciently large," and
try to give an intuitive feeling for the Central Limit Theorem. EXAMPLE Let Y be the mean of a random sample of n = 25 currents (in milliamperes) in a stripjf wire in which each measurement has a mean of 15 and a variance of 4.
Then X has an approximate N(15, 4/25) distribution. For illustration, _ 14.4 — 15 Y — 15 15.6 — 15
1 .4 . =
P( 4 < X < 15 6) P( 0.4 < 0.4 < 0.4 ) % <1>(1.5) — <1>(—1.5) = 0.9332 — 0.0668 2 0.8664. ._— EXAMPLE Let X1 , X2, . . . ,X20 denote a random sample ofsize 20 from the uniform distribution
U(0,1). Here E(X,) = 1/2 and Var(X,) = 1/12, for i = 1,2,...,20. If Y = X1+ X2 +  .  + X20. then M = 20(112) and 03 = 20(1/12) and Y — 20(1/2) < 9.1 — 10
t/20/12 ‘ ./20/12 m ¢>(—0.697) = 0.2423. P(Y 5 9.1) = P( > = P(W 5 —0.697) Also, .— Y—10 11.7—10
P(8.55Y§11.7)=P(85 10 ) (573‘ 5 m E m
= P(~l.162 : W 5 1.317)
m cb(1.317)— <1>(—1.162) = 0.9061 — 0.1226 = 0.7835.
._— EXAMPLE Let Y denote the mean of a random sample of size 25 from the distribution whose
p.d.f. isﬂx) = x3/4, 0 < x < 2. It is easy to show that it = 8/5 = 1.6 and 02 = 8/75. 144 Chapter3 Continuous Distributions EXAMPLE Thus l.5~1.6 < )7—16 < 1.65—1.6
Jam/J3 “ Jam/J2? _ Jam/m = P(—1.531 5 W 5 0.765) P(1.5 5 X 51.65): P( e waves; A <p{—1.531 ) = 0.7779 — [1.0629 = 0.7150. These examples have shown how the Central Limit Theorem can be used for
approximatingcertain probabilities concerning the mean X or the sum Y = X;
ofa random sample. That is? is approximately N(u. 02/21), and Y is approximately
N(nu,naz) when n is “sufﬁciently large," where [1 and 02 are the mean and the
variance of the underlying distribution From which the sample arose. Generally.
if n is greater than 25 or 30, these approximations will be good. However, if the
underlying distribution is symmetric, unimodal, and of the continuous type, a value
of n as small as 4 or 5 can yield a very adequate approximation. Moreover, if the
original distribution is approximately normal, X would have a distribution veryI '
close to normal when n equals 2 or 3. In fact, we know that if the sample is taken
from Nm. 0‘2), X is exactly NW, 02/") for every n = 1,13,. .. . The following examples will help to illustrate the previous remarks and will give '
the reader a better intuitive feeling about the Central LimitTheorem. In particular, _
we shall see how the size of n affects the distribution of X and Y = 2le X, for 
samples from several underlying distributions. Let X1.X2,X3,X4 be a random sample of size 4 from the uniform distribution
U(0, 1) with p.d.f. f(x) = 1.0 < x < 1. Then [1, = 1/2 and 02 = 1/12. We shall
compare the graph of the p.d.f. of
II
Y = Z X.
[:1 with the graph of the N[n(lf2),n(1f12}] p.d.f. for n = 2 and 4, respectively.
By methods given in Section 5.2 we can determine that the p.d.f. of Y = X1+X2 is % 0<yEL
go) = 2—y. 1<y<2. This is the triangular p.d.f. that is graphed in Figure 3.6—1(a). In this ﬁgure the
N[2(1/2),2(1/12)] p.d.f. is also graphed.
Moreover, the p.d.f. of Y = X) + X2 + X3 + X4 is g, 05y<L
—3y3 +12y2 _ 12y + 4
————6 .
3y3 — 24y2 + 60y ~ 44
6 .
—y3 +12y2 — 48y + 64
6 . 15y<2, 25y<l 3<y£4 EXAMPLE @ Section 3.6 The Central Limit Theorem 1.45 This p.d.f. is graphed in Figure 3.61th along with the N[4(1/2),4(1/12)] p.d.f. If
we are interesled in ﬁnding P(1.7 5 Y 5 3.2). this‘ could be done by evaluating 3.2 ‘.
f 80’) dy. ‘5
1.7 J which is tedious (see Exercise 3.68). It is much easier to use a normal approxima tion, which results in a number very close to the exact value.
_— 30’)
M2, 4/12) NU. 2/12) 1.0 ' 1.:
(a) Figure 3.6—1 p.d.f.s of sums of uniform random variables In Example 3.64 and Exercise 3.68 we show that even for a small value ofn, like
n = 4, the sum of the sample items has an approximate normal distribution. The
following example illustrates that for some underlying distributions (particularly
skewed ones) u must be quite large to obtain a satisfactory approximation. In order
to keep the scale on the horizontal axis the same for each value of n, we will use
the following result. Let ﬁx) and F(x) be the p.d.f. and distribution functiOn of a random variable,
X, of the continuous type having mean t: and variance :72. Let W = (X — #)/a.
The distribution function of W is given by G(w)=P(W 5 w) : rift“ _<_ w) =P(X 5 ow + p.) = F(aw + ,u).
Thus the p.d.f. of W is given by the derivative of F(aw + it), namely g(w) = af(aw + it). Let X1.X2,. . . ,X” be a random sample of size n from a chi—square distribution
with one degree of freedom. If 146 Chapter3 Continuous Distributions then we show (Chapter 6) that Y is X201), and E(Y) = n, Va1'(Y) = 2n. Let Y—n
W: .
«32:1 The p.d.f. of W is given by /2 n/Z—l
g(w) = v2n (—LHL— e_(‘/§;"+”m, —n/V2n < w < oo. row Note that w > —n/\/2n corresponds to y > 0. In Figure 3.62(a) and (b), the graph
of W is given along with the N“). I) p.d.f. for n = 20 and 100, respectively. ___._.__ Figure 3.62 p.d.f.s of W = (Y  lzlfm} So far all the illustrations have concerned distributions ol the continuous type.
However, the hypotheses for the Central Limit Theorem do not require the
distribution to be continuous. We shall. consider applications DI the Central Limit
Theorem for discretetype distributions in the next section. ESEEHCISESEG 3.61 Let 3? be the mean of a random sample oisize 12 from the uniform distribution on
the interval (0,1). Approximate P(1/2 g X g 2/3). 3.62 Let l" 2 X] + X2 +    + X15 be the sum of a random sample of size 15 from the
distribution whose p.d.f. isﬂx) = (3/2))(2, —1 < x < 1. Approximate P(—0.3 5 Y 515). 3.63 Let 7 be the mean of a random sanEle of size 36 from an exponential distribution
with mean 3. ApprOxirrrate P(2.5 5 X 5 4}. 3.64 A random sample ol size n = [E is taken from the distribution with p.d.f f(x) :
l—x/2,05x 52. (3) Find a and 02. (b) Find, approximately. P(2/3 s f 5 536). 3.65 Let X equal the maximal oxygen intake 0! a human on a treadmill, where the
measurements are in milliliters of oxygen per minute per kilogram of weight. Section 3.6 The Central Limit Theorem 1.47 Assume that for a particular population the mean of X is it = 54.030 and the
standard deviation tScr 22.8. Let X be the sample mean of a random sample of size
n = 47. Find P(52.761 5 it 5 54.453), approximately. 3.66 Let X equal the weight in grams of a miniature candy bar. Assume that it = E(X) = 24.43 and 0'2 = Var(X) = 2.20. Let 7 be the sample mean of a random sample of
H .= 30 candy bars. Find (.3 E(X). (b) waif). (:1 £424.17 5 X 5 24.82), approximately. 3.67 Let X equal the birth weight in grams; of a baby born in the Sudan. Assume that
E(X) = 3320 and VarUf] z Let X be the sample mean of a random sample of
size n = 225. Find H3233.76 5 X 5 3406.24}, approximately. 1n Example 3.64. with n = 4. compute P{l.7 5 Y 5 3.2) and compare this answer
with the normal approximation of this probability. Five measurements in "ohms per square“ of the electronic conductive coating (that
allows light to pass through it) on a thin, clear piece of glass were made and the
average was calculated. This was repeated 150 times. yielding the following averages: 83.36 75.86 79.65 90.57 95.37 97.97 77.00 80.80 83.53 83.17
80.20 31.42 89.14 88.68 85.15 90.11 89.03 85.00 82.57 79.46
81.20 82.80 81.28 74.64 69.85 75.60 76.60 78.30 88.51 86.32
84.03 95.27 90.05 77.50 75.14 81.33 86.12 78.30 80.30 84.96
79.30 88.96 82.76 83.13 79.60 86.20 85.16 87.86 91.00 91.10
84.04 92.10 79.20 83.76 87.87 86.71 81.89 85.72 75.84 74.27
93.08 81.75 75.66 75.35 76.55 84.86 90.68 91.02 90.97 98.30
91.84 97.4] 73.60 90.65 80.20 74.75 90.35 79.66 86.88 83.00
86.24 80.50 74.25 91.20 70.16 78.40 85.60 80.82 75.95 80.75
81.86 8218 82.98 84.00 76.85 85.00 79.50 86.56 83.30 72.40
79.20 86.20 82.36 84.13 86.11 88.25 88.93 93.12 78.30 77.24
82.52 31.37 83.72 86.90 84.37 92.60 95.01 78.95 81.40 88.40
76.10 85.33 82.95 80.20 88.21 83.49 81.00 82.72 81.12 83.62
91.18 85.90 79.01 77.56 81.13 80.60 81.65 70.70 69.36 79.09
71.35 67.20 67.43 69.95 66.76 76.35 69.45 80.13 84.26 88.13 (a) Construct a frequency table for these 150 observations using 10 intervals of
equal length, and using {66.495. 69.695} for the ﬁrst class interval. Which is the
modal class? (b) Construct a relative frequency histogram for the grouped data. (c) Superimpose a normal p.d.f. using the mean 82.661 and variance 42.2279 of
these 150 values as the true mean and variance. Do these 150 means of ﬁve
measurements seem to be normally distributed? A church has pledges (in dollars) with a mean of 2000 and a standard deviation of
500. A random sample of size n = 25 is taken from all of this church’s pledges and the sample mean X is considered. Approximate NY > 2050). The tensile strength X of paper has it = 30 and a = 3 (pounds per square inch). A
random sample of size n =: 100 is taken from the distribution of tensile strengths.
Compute the probability that the sample mean T is greater than 29.5 pounds per
square inch. At certain times during the year, a bus company runs a special van holding ten
passengers from Iowa City to Chicago. After the opening of sales of the tickets, the
time {in minutes) between sales of tickets for the trip has a gamma distribution with
a := 3 and a = 2. Approximate the probability of being sold out within one hour. Let Kink}, X3. X4 represent the random times in days needed to complete four
steps of a project. These times are independent and have gamma distributions with
common 6 = 2 and common a = 4. One step must be completed before the next can
be started. Let Y equal the total time needed toeomplete the project. Approximate
P‘(Y 5 35). Does it seem appropriate to use the Central Limit Theorem when n = 4
in this exercise? 148 Chapter3 Continuous Distributions APPROXIMA
TIONS FOR
DISCRETE
DISTRIBU
TIONS EXAMPLE 3.614 Assume that the sick leave taken by the typical Worker per year has it = 10, a = 2,
measured in days A ﬁrm has n = 20 employees. Assuming independence, how
many sick days should the ﬁrm budget if the ﬁnancial ofﬁcer wants the probability
of exceeding the budgeted days to be less than 20%? In this section we illustrate how the normal distribution can be used to approximate
probabilities for certain discretetype distributions. One of the more important
discrete distributions is the binomial distribution. To see how the Central Limit
Theorem can be applied, recall that a binomial random variable can be described
as the sum of Bernoulli random variables. That is, let X1,X2. . . . ,X,, be a random
sample from a Bernoulli distribution with a mean u = p and a variance 02 :—
p(1 —p), where 0 < p < 1. Then Y = X, is b(n.p). The Central Limit Theorem states that the distribution of Y—np : X—p
W W is N(0, 1) in the limit as n —> 00. Thus, if n is sufﬁcientiy large, the distribution of Y
is approximately N[np,np(l —p]]. and probabilities for the binomial distribution
b(n,p) can be approximated using this normal distribution. A rule often stated is
that n is “sufﬁciently large” it np =_ 5 and at] —p) 3 5. This can be used as a rough
guide although as p deviates more and more from 0.5, we need larger and larger
sample sizes, because the underlying Bernoulli distribution becomes more skewed.
Note that we shall be approximating probabilities from a discrete distribu
tion with probabilities for a continuous distribution. Let us discuss a reasonable
procedure in this situation. Let Y be b(n,p). Recall that the probability W: l , n' k II—k
P{l’ =k)::f[it] = mp (1 —p) , k = 0,1,2,...,n
can be represented graphically as the area of a rectangle with a base of length one
centered at k (that is. the base goes from k — U2 to k + NZ} and a height equal to
P(X 2 k). The collection of these rectangles for k = t], 1.. ...n is the probability
histogram For Y. Figures 3.7l(a} and (b) show the graphs of the probability
histograms for the MID. 1,12} and btlS, 1/6) distributions, respectively. When using
the normal distribution to approximate probabilities for the binomial distribution.
areas under the p.d.f. for the normal distribution will be used to approximate areas
of rectangles in the probability histogram for the binomial distribution. That is, the
area under the Nianlpﬂ w p)] p.d.f. between k — 1;? and k + 1/2 will be used to
approximate P(Y = k). Let Y be b(10, 1/2). Then, using the Central Limit Theorem, P(a < 1" < b} can be
approximated using the normal distribution with mean 10(1f2) = 5 and variance
10(1/2)(1/2) = 5/2. Figure 3.71(a) shows the graph of the probability histogram
for b(10,1f2) and the graph of the p.d.f. oi the normal distribution N(5, 5/2). Note
that the area of the rectangle whose base is 1 l
k——.k —
i 2 +2) and the area under the normal curve between It —1;‘2 and k+ 1/2 are approximately equal for each integer k.
_— Appendix 0 ' Answers to OddNumbered Exercises 349 3.4? (a) *'
Stems Leaves Frequencies Depths
III 3 1 l
121 U 3 2 3
12 5 6 2 5
13* l 3 4 3 S
13. S 5 7' if T 9 E 14
14* 0012 3 4 4 4 B 22
141 5 E 7 'l" 7 8 9'9 3 30
15* 1100011123344 12 3U
15. 5 5 67 3 8 8 9 S 13
16* l] (l U 2 3 4 6 10'
IE. 5 5 2
17* l l 2
17 5 1 1
3.49 0.025.
3.4! l 0.514.
3.41.1 c =1/W.
3.5] (31 = r2 = 33.4267; 02 = 33 = 5.0980. 3.5.3 (0) (i) 0 = 0.54930 = 0.5975.
(ii) (3 = 2.21013 = 2.4004,
(iii) a“ = 0.958813 = 0.8646. 3.5.5 0: min(X1,X2,...,X,,).
_ _ __ x x
3.37 [x(1—2/ﬁ},x{l+2/Jﬁ]] or
3.0] 0.4772.
3.03 0.8185.
3.65 0.6247.
3.07 0.95. 3.09 (a) The frequencies are 5, 6, 12, 18, 31, 31, 20, 17, 6, 4; the ﬁfth and sixth classes; (1:) yes.
3.6! 1 0.9522. $1113 P(Y 5 35) 8 0.6462 using normal approximation; P(Y 5 35) = 0.6725 using the gamma distribution
with 01 = 16 and (9 = 2. 3.7l (a) 0.2878, 0.2881; (b) 0.4428, 0.4435; (6:) 0.1550, 0.1554.
3.73 0.9258 using normal approximation, 0.9258 using binomial. ...
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 Fall '07
 Guggenberger
 Central Limit Theorem, Normal Distribution, Probability theory, p.d.f.

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