142-148 Chapter 3.6 The Central Limit Theorem

142-148 Chapter 3.6 The Central Limit Theorem - 142...

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Unformatted text preview: 142 Chapter3 Continuous Distributions THE CENTRAL LIMIT THEOREM 3.5-7 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 3-6-1 [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 3.6-1th 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 3.6-8). 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 3.6-8 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 3.6-2(a) and (b), the graph of W is given along with the N“). I) p.d.f. for n = 20 and 100, respectively. ___._.__ Figure 3.6-2 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 discrete-type distributions in the next section. ESEEHCISESEG 3.6-1 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). 3.6-2 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). 3.6-3 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}. 3.6-4 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). 3.6-5 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. 3.6-6 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. 3.6-7 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 3.6-4. 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 3.6-14 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 discrete-type 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 3.7-l(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 3.7-1(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 Odd-Numbered 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 3.4-9 0.025. 3.4-! l 0.514. 3.4-1.1 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 3.3-7 [x(1—2/fi},x{l+2/Jfi]] or 3.0-] 0.4772. 3.0-3 0.8185. 3.6-5 0.6247. 3.0-7 0.95. 3.0-9 (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. $11-13 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. 3.7-l (a) 0.2878, 0.2881; (b) 0.4428, 0.4435; (6:) 0.1550, 0.1554. 3.7-3 0.9258 using normal approximation, 0.9258 using binomial. ...
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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.

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142-148 Chapter 3.6 The Central Limit Theorem - 142...

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