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STAT 330 Spring 08 Assignment 3 Dr. Paula Smith Due Thurs July 10 1. (Adapted from Larson, Intro. to Probability Theory, 1969) A nursery specializes in the installation of circular flower beds. When laying out the circle, a workman puts a peg in the center and cuts a length of rope (already tied in a loose loop to the stake) equal to the radius of the desired circle, and uses this to mark out the bed on the ground. Assume the desired radius is r meters. Also assume the workman is a little sloppy, and is equally likely to cut the rope to any length within the interval (r – 0.1, r + 0.1). Let X be the length of the rope from stake to cut end. a. Determine the probability density function for X. b. Let Y be the surface area of the circle. State Y in terms of X. c. Determine the probability density function for Y. d. Find E(Y) e. Find P(Y > r2). Solution a) Since it is assumed that the workman is a little sloppy, and is equally likely to cut the rope to any length within the interval (r – 0.1, r + 0.1), the random variable X, the length of the rope from stake to cut end follows a Uniform distribution over the range (r – 0.1, r + 0.1). So the probability density function of X is given by 1 f X ( x) , r 0.1 x r 0.1 0.2 = 0 otherwise [Note that the probability density function of a Uniform distribution over the range (a ,b) is 1 f X ( x) , a xb] ba b) Let Y be the surface area of the circle. Then Y can be written as Y X 2 [Note that the surface area of a circle with radius r is r 2 ] c) Since X varies from (r – 0.1) to (r + 0.1), Y varies from (r 0.1)2 to (r 0.1)2 Now the probability density function of Y is given by dx gY ( y ) f X ( x) dy From the transformation y x 2 we get x y and hence dx 1 . dy 2 y Thus the probability density function of Y is given by 1 1 gY ( y) 0.2 2 y 1 , (r 0.1)2 y (r 0.1)2 , 0.4 y = 0 otherwise [Note that the above method to find the p.d.f. is valid because the transformation from X to Y is one to one as X takes only positive values since r >0.1 always.] d) we have E (Y ) ygY ( y )dy y ( r 0.1)2 y ( r 0.1)2 1 0.4 1 0.4 1 0.6 1 0.4 y dy ( r 0.1)2 ydy ( r 0.1)2 ( r 0.1)2 2 3/ 2 3 y ( r 0.1)2 (r 0.1)2 3/ 2 (r 0.1) 2 3/ 2 (r 0.1)3 (r 0.1)3 0.6 3 2 (r 0.3r 0.03r 0.001) (r 3 0.3r 2 0.03r 0.001) 0.6 0.6 (0.6r 2 0.002) e) We have P[Y r ] ( r 0.1)2 2 r gY ( y )dy 2 ( r 0.1)2 r 2 1 0.4 1 0.4 1 0.2 1 0.4 y ( r 0.1)2 r2 dy 1 dy y ( r 0.1)2 2 y 2 r (r 0.1)2 r 2 (r 0.1) r 0.2 0.1 0.2 1 = = 0.50 2 2. The Weibull density function is given by f(y) = (1/) m ym–1 exp(–ym /) if 0< y <; and f(y) = 0 otherwise, where and m are positive constants. This density function is often used as a model for the length of life of physical systems. Suppose Y has the Weibull density just given. a. Find the density function of U = Ym. b. Find E(Yk) for any positive integer k. c. Suppose W has an exponential distribution with mean . Prove that Y = W has a Weibull density with = and m = 2. d. Find E(Wk/2) Solution Here it is given that the p.d.f. of Y is ym 1 f ( y) m y m1 exp , 0 y a) Let U Y m . Let g (u ) be the probability density function of U. Then we have g (u ) f ( y ) dy du Since u y m , we have y u1/ m and hence dy u (1/ m )1 du m Thus the probability density function of U is given by 1 g (u) m u m 1 m (1/ m ) 1 uu exp m 1 u exp , 0 u Hence U Y m follows an Exponential distribution with mean . b) E (Y k ) y k f ( y )dy 0 ym 1 k m1 m y y exp dy 0 ym 1 m y k m1 exp dy 0 Put u y m , then y u1/ m and hence dy u (1/ m )1 du m Now the above integral becomes 1 E (Y k ) m u 0 k m 1 m (1/ m ) 1 u u exp du m k 1 u u m exp du 0 k 1 11 u u m exp du 0 k 1 k k 1 m , where 1 Gamma 1 k 1 m m 1 m k k / m 1 m c) Suppose W has an exponential distribution with mean . Then the density function of W is given by w 1 h( w) exp , 0 w Consider the transformation Y W . Now the p.d.f. of Y is given by f ( y ) h( w) dw dy Since y w , we have w y 2 and hence dw 2y dy Thus the p.d.f. of Y is given by y2 f ( y) exp 2 y y2 1 2 y 21exp , 0 y , which is the p.d.f. of a Weibull density with = and m = 2. 1 d) Since Y = W has a Weibull density with = and m = 2, using the result (b) we have E (W k / 2 ) E (Y k ) k k / 2 1 2 3. A small orchard contains 5 Winesap apple trees and 3 Mutsu apple trees, all at peak bearing age. The yearly harvests for Winesap apple trees at that stage are normally distributed with mean 1 and variance 2, while the yearly harvest for Mutsu apple trees at that stage are normally distributed with mean 2 and variance 22. a. What is the expected yearly harvest for this orchard? b. Assume you have lists of yearly harvest amounts per tree and you can randomly select n of the Winesap harvest amounts and m of the Mutsu harvest amounts. Suppose 2 is known. Construct a 95% confidence interval for the expected yearly harvest for this orchard. c. Under the same conditions as in part b, but with 2 unknown, construct a 95% confidence interval for the expected yearly harvest for this orchard. Solution Let X denotes the yearly harvests for Winesap apple trees and Y denotes the yearly harvest for Mutsu apple trees in the orchard. Here it is given that X follows a Normal distribution with mean 1 and variance 2 and Y follows a Normal distribution with mean 2 and variance 22. a) Since the orchard contains 5 Winesap apple trees and 3 Mutsu apple trees, the expected yearly harvest for this orchard is E(5X + 3Y) = 5E(X) + 3E(Y) = 51 32 . b) Let U= 5X + 3Y. Then U represents the yearly harvest for the orchard. Since X follows a Normal distribution with mean 1 and variance 2 and Y follows a Normal distribution with mean 2 and variance 22, assuming that X and Y are independent U= 5X + 3Y follows a Normal distribution with mean 51 32 and variance 432. [Note that Var(U) = Var(5X + 3Y)=25Var(X)+9Var(Y)=252 + 9*22 = 432] Let x1 , x2 ,..., xn denote the random sample of the Winesap harvest amounts and y1 , y2 ,..., ym denote the random sample of the Mutsu harvest amounts. Suppose x y 1 n xi and n i 1 1 m yi are the sample means of the two random samples respectively. m i 1 Since x1 , x2 ,..., xn are random samples from a Normal distribution with mean 1 and variance 2, x follows a normal distribution with mean 1 and variance x 2 . That is n N ( 1 , 2 n) . Similarly since y1 , y2 ,..., ym are random samples from a Normal distribution with mean 2 and variance 22, y follows a normal distribution with mean 2 and variance 2 2 . That is y N ( 2 , 2 2 m) . m Thus 5 x 3 y follows a normal distribution with mean 51 32 and variance 25 2 n 9 2 2 25 18 2 . m n m 25 18 N (51 32 , 2 ) n m 5 x 3 y (51 32 ) Hence Z N (0,1) , a Standard Normal distribution. 25 18 n m For a Standard Normal random variable Z, we know that That is 5 x 3 y P[1.96 Z 1.96] 0.95 5 x 3 y (51 32 ) That is P 1.96 1.96 0.95 25 18 n m 25 18 25 18 i.e., P 1.96 5 x 3 y (51 32 ) 1.96 0.95 n m n m 25 18 25 18 i.e., P (5 x 3 y ) 1.96 (51 32 ) (5 x 3 y ) 1.96 0.95 n m n m Thus a 95% confidence interval for the expected yearly harvest for this orchard is 25 18 25 18 (5 x 3 y ) 1.96 , (5 x 3 y ) 1.96 n m n m c) Let S X2 1 m 1 n 2 2 ( x x ) S ( yi y ) 2 denote the sample variances of the two and Y i m i 1 n i 1 samples. Then we have Also 2 Y 2 mS 2 nS X2 (2n 1) , a Chi-Square distribution with n 1 degrees of freedom. 2 mSY2 2 2 2 5 x 3 y (51 32 ) 2 ( m 1) . Thus nS X2 (2n m2) 25 18 Hence t n m nS X2 mSY2 2 (n m 2) 2 2 5 x 3 y (51 32 ) 25 18 mSY2 2 (n m 2) n m 5 x 3 y (51 32 ) t( n m2) , a Student’s t distribution with (n m 2) degrees of 25 18 SP n m nS 2 X freedom, where SP [Note that if X nS 2 X mSY2 2 (n m 2) is the pooled standard deviation. N (0,1) and Y (2n) , then t X Y n t( n ) ] From Student’s t –table with (n m 2) degrees of freedom, we can find a t0.025 such that P[t0.025 t t0.025 ] 0.95 5 x 3 y (51 32 ) That is P t0.025 t0.025 0.95 25 18 SP n m 25 18 25 18 i.e., P t0.025 S P 5 x 3 y (51 32 ) t0.025 S P 0.95 n m n m 25 18 25 18 i.e., P (5 x 3 y ) t0.025 S P (51 32 ) (5 x 3 y ) t0.025 S P 0.95 n m n m Thus a 95% confidence interval for the expected yearly harvest for this orchard is 25 18 25 18 (5 x 3 y ) t0.025 S P , (5 x 3 y ) t0.025 S P n m n m