chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions79.1=200p ; n = 1000; λ = np = 5a. P(5 ≤ X ≤ 8) = F(8;5) – F(4;5) = .492b. P(X ≥ 8) = 1 – P(X ≤ 7) = 1 - .867 = .13380.a. The experiment is binomial with n = 10,000 and p = .001,so µ = np = 10 and σ = npq = 3. 161 .b. X has approximately a Poisson distribution with λ = 10,so P(X > 10) ˜ 1 – F(10;10) = 1 - .583 = .417c. P(X = 0) ˜ 081.a. λ = 8 when t = 1, so P(X = 6) = F(6;8) – F(5;8) =.313 - .191 = .122,P(X ≥ 6) = 1 - F(5;8) = .809, and P(X ≥ 10) = 1 - F(9;8) = .283b. t = 90 min = 1.5 hours, so λ = 12; thus the expected number of arrivals is 12 and the SD= 12 = 3. 464c. t = 2.5 hours implies that λ = 20; in this case, P(X ≥ 20) = 1 – F(19;20) = .530 and P(X ≤10) = F(10;20) = .011.82.a. P(X = 4) = F(4;5) – F(3;5) = .440 - .265 = .175b. P(X ≥ 4) = 1 - P(X ≤ 3) = 1 - .265 = .735c. Arrivals occur at the rate of 5 per hour, so for a 45 minute period the rate is λ = (5)(.75)= 3.75, which is also the expected number of arrivals in a 45 minute period.83.a. For a two hour period the parameter of the distribution is λt = (4)(2) = 8,so P(X = 10) = F(10;8) – F(9;8) = .099.b. For a 30 minute period, λt = (4)(.5) = 2, so P(X = 0) = F(0,2) = .135c. E(X) = λt = 2119