, find P(X = 2 or 3) et) 5. )px (1 p) n x x = 0, 1, 2,..., n. 0 elsewhere = 40

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1 Assignment 4 Fall 07. Exercise 3.. on Page 46: If the mgf of a rom variable X is ( et) 5, find P(X or 3). Since the M(t) of X is ( et) 5, X has a binomial distribution with n 5, p 3. The probability density function of the binomial distribution is ( n x p(x) )px ( p) n x x 0,,,..., n 0 elsewhere Thus, P(X or X 3) P(X ) + P(X 3) ( ) ( ) 5 ( ) 3 ( ) ( ) 5 3 ( ) Exercise 3..7 on Page 49: Consider a shipment of 000 items into a factory. Suppose the factory can tolerate about 5% defective items. Let X be the number of defective items in a sample without replacement of size n 0. Suppose the factory returns the shipment if X. (a) Obtain the probability that the factory returns a shipment of items which has 5% defective items. (b) Suppose the shipment has 0% defective items. Obtain the probability that the factory returns such a shipment. (c) Obtain approximations to the probabilities in parts (a) (b) using appropriate binomial distributions. Note : If you do not have access to a computer package with a hypergeometric comm, obtain the answer to (c) only. This is what would have been done in practice 0 years ago. If you have access to R, then the comm dhyper(x, D, N D, n) returns the probability in expression (3..7) Stats3D03

2 Assignment 4 Fall 07 The expression (3..7) is p(x) (N D n x )(D x ) ( N n ), x 0,,..., n In the following, let X be the number of defective items. (a) From the given information, we have that N 000, n 0, D 000 5% 50 Since X has a hyper-geometric distribution, we have that P(X ) P(X 0) P(X ) ( )(50 0 ) (b) Since D 000 0% 00, we have that ( ) ( )(50 ) ( ) P(X ) P(X 0) P(X ) ( )( 00 0 ) ( ) ( )( 00 ) ( ) (c) From the given information, when n 0, p 0.05, using the binomial distributions, we have that P(X ) P(X 0) P(X ) ( ) ( 0.05) ( ) ( 0.05) 0 When n 0, p 0., using the binomial distributions, we have that P(X ) P(X 0) P(X ) ( ) ( 0.) ( ) 0 0. ( 0.) 0 3. Exercise 3..3 on Page 54: In a lengthy manuscript, it is discovered that only 3.5 percent of the pages contain no typing errors. If we assume that the number of errors per page is a rom variable with a Poisson distribution, find the percentage of pages that have exactly one error. Stats3D03

3 Assignment 4 Fall 07 Since the Poisson distribution is P(X x) e λ λ x x!, x 0,, we have P(X 0) e λ λ 0 From the given information, P(X 0) 3.5%. Therefore, we can obtain λ. Thus, 0! 4. Exercise 3..4 on Page 55: P(X ) e λ λ! e! 0.7. Let X X be two independent rom variables. Suppose that X Y X + X have Poisson distributions with means µ µ > µ, respectively. Find the distribution of X The mgfs of X Y that have Poisson distributions are given by, respectively M X (t) exp{(µ (e t )} M Y (t) exp{(µ(e t )} Since X X are independently rom variables, we have M Y (t) M X (t)m X (t) exp{(µ(e t )} exp{(µ (e t )}M X (t) Therefore, M X (t) exp{(µ µ )(e t )} Thus, X has poisson distribution with mean (µ µ ). Stats3D03 3

4 Assignment 4 Fall Exercise on Pages 64 Let X, X, X 3 be iid rom variables, each with pdf f (x) e x, 0 < x <, zero elsewhere. (a) Find the distribution of Y minimum(x, X, X 3 ). Hint : P(Y y) P(Y > y) P(Xi > y, i,, 3). (b) Find the distribution of Y maximum(x, X, X 3 ). From the given information, we have that the cumulative density function is F(x) e x (a) F Y (y) P(Y y) P(Y > y) P(X > y, X > y, X 3 > y) [ P(X y)][ P(X y)][ P(X 3 y)] [ ( e y )] 3 e 3y Thus the distribution of Y is e 3y y > 0 F Y (y) 0 elsewhere (b) F Y (y) P(Y y) P(X y)p(x y)p(x 3 y) ( e y )( e y )( e y ) ( e y ) 3 Thus the distribution of Y is ( e y ) 3 y > 0 F Y (y) 0 elsewhere Stats3D03 4

5 Assignment 4 Fall Exercise on Page 76: Let X be a rom variable such that E(X m ) (m)!/( m m!), m,, 3,... E(X m ) 0, m,, 3,.... Find the mgf the pdf of X. The moment generating function of X is M X (t) Ee (tx) E [ n0 ] (tx) n n! E(X n )t n n0 n! m m0 + (t /) m m! E(X m )t m (m)! (Since EX m 0) e t Thus, X follows N(0, ), the pdf is f (x) π exp 7. Exercise 3.6. on Page 96: Show that Y ) ( x, < x < + (r /r )W where W has an F -distribution with parameters r r, has a beta distribution. Let U V are independent chi-square rom variables with r r degrees of freedom, respectively, let W r U/(r V). Then we can obtain that W has an F-distribution with two parameters r r. Y + (r /r )W + (r /r )( r U r V ) V U + V Stats3D03 5

6 Assignment 4 Fall 07 Since U V have chi-square distribution with r r degrees of freedom, we obtain that V U+V has a beta distribution. Thus Y has a beta distribution. 8. Exercise on Page 96: Let X, X, X 3 be three independent chi-square variables with r, r, r 3 degrees of freedom, respectively. (a) Show that Y X /X Y X + X are independent that Y is χ (r + r ). (b) Deduce that are independent F-variables. X /r X /r (X + X )/(r + r ) (a) Since Y X /X Y X + X, we can obtain that X Y Y /( + Y ) X Y /( + Y ). The determinant of Jacobian is Therefore the joint pdf of X X is y (y +) y y + y (y +) y + y (y + ) f (x, x ) r +r Γ ( r ) Γ ( r ) x r r x e x+x The joint pdf of Y Y is f (y, y ) ( ) y r +r Γ ( y r ( ) r y r ) ( Γ r ) e y y y + y + (y + ) r +r Γ ( r ) ( Γ r ) y r y r +r (y + ) r +r e y Thus the joint pdf can be decomposed into two factors, each of them relevant to Y, Y alone. Hence Y, Y are independent. Since X X follow independently chi-square distribution with r, r, we have that Y has a chi-square distribution with r + r. Stats3D03 6

7 Assignment 4 Fall 07 (b) Since X, X, X 3 are three independent chi-square variables with r, r, r 3 degrees of freedom, therefore W X /r X /r F(r, r ) Y X + X χ (r + r ) From the known information, the ratio of two chi-square variables with their respective degrees of freedom follows F-distribution. Let V Y /(r + r ) (X + X )/(r + r ) F(r 3, (r + r )) X, X, X 3 are independent, X /X, Y are independent. So W V are independent. Thus are independent F-variables. X /r X /r (X + X )/(r + r ) Stats3D03 7