1. Section 3.5 Let X have a gamma( ,  ) with  = r/2, where r is a positive integer, and  = 2. We say that X has a chi-square distribution with r degrees of freedom (df), denoted  2 (r). The family of chi-square distributions has special significance which becomes apparent with the study of some other distributions (such as the normal distribution). (Table IV in Appendix B of the textbook provides certain values of the distribution function for selected  2 (r) random variables and is designed especially for applications to be discussed later.) The p.d.f. of X is f(x) = E(X) =Var(X) = The m.g.f. of X is x r/2 – 1 e –x/2 ———— if x > 0  (r/2) 2 r/2 r2r2r 1 ——— for t < 1 / 2 (1 – 2t) r/2 M(t) =

2. 1. (a) Flaws in a certain type of recording tape follow a Poisson distribution and occur at an average rate of 3 flaws per 5000 ft. The following random variables are defined: X = number of thousands of feet until the first flaw is observed, Y = number of thousands of feet until the second flaw is observed, V = number of thousands of feet until the 6th flaw is observed. Identify the type of distribution that each of X, Y, and V has. X has an distribution Y has adistribution. V has a distribution. exponential(5/3) (a gamma(1,5/3) distribution). gamma(2,5/3) gamma(6,5/3)

3. (b)Find each of the following: P(X > 2) 2  3e – 3x / 5 ——— dx= 5 – e – 3x / 5 x = 2  = e – 6 / 5 = 0.301 or Define the random variable W = number of flaws in 2 thousand feet. W has a Poisson(6/5) distribution. P(X > 2) = P(W = 0)= 0.301 (from the Poisson Distribution Table)

4. P(Y < 4) 0 4 9y e – 3y / 5 ———— dyor 25 Define the random variable W = number of flaws in 4 thousand feet. W has a Poisson(12/5) distribution. P(Y < 4) = P(W  2) = 1 – P(W  1) = 0.692 (from the Poisson Distribution Table) 0 9 729v 5 e – 3v / 5 ————— dv or 1875000 Define the random variable W = number of flaws in 9 thousand feet. W has a Poisson(27/5) distribution. P(V < 9) = P(W  6) = 1 – P(W  5) = 0.454 (from the Poisson Distribution Table) P(V < 9)

5. 2. (a) Flaws in a certain type of wire follow a Poisson distribution and occur at the average rate of 5 per thousand feet. Note that in order to use the chi-square tables to obtain probabilities concerning a gamma random variable, we must have  = 1/ = 2, that is, = 1/2. Since flaws occur at an average rate of 5 per thousand feet, we can also say that flaws occur at an average rate of 1/2 = 0.5 per hundred feet. From the chi-square tables, what can be said about the probability that more than 1244 feet of wire will be used before the 10th flaw, Define the random variable X = number of hundreds of feet until the 10th flaw is observed. X has a gamma(10,2) distribution, which is a  2 ( ) distribution. 20 P(X > ) =12.44 1 – P(X  12.44) = 0.90

6. P(X > ) =31.41 1 – P(X  31.41) = 0.05 (b) (c) more than 3141 feet of wire will be used before the 10th flaw, Using the random variable X defined in part (a), we have less than 1831 feet of wire will be used before the 5th flaw, 10 P(Y < ) =18.31 P(Y  18.31) = 0.95 Define the random variable Y = number of hundreds of feet until the 5th flaw is observed. Y has a gamma(5,2) distribution, which is a  2 ( ) distribution.

7. Define the random variable W = number of hundreds of feet until the 2nd flaw is observed. W has a gamma(2,2) distribution, which is a  2 ( ) distribution. 4 P( < W < ) =7.7799.488 P(W  9.488) – P(W  7.779) = 0.95 – 0.90 = 0.05 (d) (e) between 777.9 and 948.8 feet of wire will be used before the 2nd flaw, more than 1900 feet of wire will be used before the 4th flaw, Define the random variable V = number of hundreds of feet until the 4th flaw is observed. V has a gamma(4,2) distribution, which is a  2 ( ) distribution. 8 P(V > ) =19 1 – P(V  19) = between 0.01 and 0.025

8. Define the random variable Q = number of hundreds of feet until the 9th flaw is observed. Q has a gamma(9,2) distribution, which is a  2 ( ) distribution. 18 P(Q < ) =36 P(Q  36) = greater than 0.99 (f)less than 3600 feet of wire will be used before the 9th flaw,