1. Section 2.6 Consider a random variable X = the number of occurrences in a “unit” interval. Let = E(X) = expected number of occurrences in a “unit” interval. Divide the unit interval into n equal subintervals. The expected number of occurrences in a subinterval is /n. (1/6) (phone calls)(one hour) (10)(phone calls)(one hour) (60) (minutes)

2. For sufficiently large n, we make the following assertions: (1) The probability of more than one occurrence in a subinterval is  0. (2) The probability of exactly one occurrence in a subinterval is  /n. (3) The probability of no occurrence in a subinterval is  1 – /n. (4) The subintervals can be treated as independent Bernoulli trials with regard to whether or not there is an occurrence. (more than one phone call) (one phone call) (1/6) (no phone call) (5/6) (a phone call)

3. Treating the subintervals as independent Bernoulli trials, we have that for x = 0, 1, 2, …, n,P(X = x) = nxnx — n x 1 – — n n – x Lim P(X = x) = n  Lim= n  nxnx — n x 1 – — n n – x Lim = n  x — x! n! ———— (n – x)! n x 1 – — n – x 1 – — n n x — (1) e – (1 – 0) = x! x e – —— x!

4. The random variable X, defined to be the number of occurrences in a unit interval assuming the four assertions are true, is said to have a Poisson distribution with mean, denoted Poisson( ). The p.m.f. of X is f(x) = E(X) = Var(X) = The m.g.f. of X is if x = 0, 1, 2, 3, … x e – —— x!

5. 1. A random variable X has a Poisson distribution with mean. Find the m.g.f. for X, and use the m.g.f. to find E(X) and Var(X). M(t) = E(e tX ) =   e tx x = 0 x e – —— = x!   x = 0 ( e t ) x e – ———– = x!   x = 0 e – ( e t ) x ——– = x! e – e = etet e for –  < t <  (e t –1) R(t) = ln[M(t)] = (e t – 1) R / (t) = e t R // (t) = e t E(X) =  = R / (0) = Var(X) =  2 = R // (0) =

6. The random variable X, defined to be the number of occurrences in a unit interval assuming the four assertions are true, is said to have a Poisson distribution with mean, denoted Poisson( ). The p.m.f. of X is f(x) = E(X) = Var(X) = The m.g.f. of X is if x = 0, 1, 2, 3, … x e – —— x! M(t) = e for –  < t <  (e t –1)

7. 2. (a) Flaws in a certain type of recording tape occur at an average rate of 3 flaws per 1500 ft. and follow a Poisson distribution. The following random variables are defined: X = the number of flaws in 1500 feet of tape, Y = the number of flaws in 3000 feet of tape. Find each of the following: the p.m.f. of X E(X) Var(X) f 1 (x) = 3 x e –3 ——if x = 0, 1, 2, 3, … x! = 3

8. (b) (c) Find each of the following: the p.m.f. of Y E(X) Var(X) f 2 (y) = Find the probability that there are no flaws in 1500 feet of tape, = 6 6 y e –6 ——if y = 0, 1, 2, 3, … y! P(X = 0) =0.05

9. are exactly two flaws in 1500 feet of tape, is more than one flaw in 1500 feet of tape, are no flaws in 3000 feet of tape, P(X = 2) = P(X  2) – P(X  1) = 0.423 – 0.199 =0.224 P(X > 1) = 1 – P(X  1) = 1 – 0.199 =0.801 P(Y = 0) =0.002

10. are exactly three flaws in 750 feet of tape, are at most five flaws in 2700 feet of tape. 3 flaws per 1500 feet = flaws per 750 feet = 1.50.934 – 0.809 = 0.125 3 flaws per 1500 feet = flaws per 2700 feet = 5.40.546

11. 3.Of all the fuses produced by a certain manufacturer, 2.5% are defective. Find the probability that a box of 200 fuses contains no more than two defects. X = the number of defects in 200 randomly selected fuses X has a distribution. E(X) = (200)(0.025) = 5 We can approximate the distribution of X with a Poisson(5) distribution, and find that P(X  2) = 0.125 b(200, 0.025)