chapter 1

Chapter 4: Continuous Random Variables and Probability Distributionsc. We want a value of X for which F(X;4)=.99. In table A.4, we see F(10;4)=.990. So withβ = 6, the 99 th percentile = 6(10)=60.d. We want a value of X for which F(X;4)=.995. In the table, F(11;4)=.995, so t =6(11)=66. At 66 weeks, only .5% of all transistors would still be operating.59.1a. E(X) = = 1λ1 = =λb. σ 1−(1)(4)−4c. P(X ≤ 4 ) = 1−e = 1 − e = . 982−(1)(5)−(1)(2)−2−5d. P(2 ≤ X ≤ 5) = 1−e − [ 1−e ] = e − e = . 12960.a.−(100)(.01386)−1.386P(X ≤ 100 ) = 1−e = 1−e = . 7499−(200)(.01386)−2.772P(X ≤ 200 ) = 1−e = 1−e = . 9375P(100 ≤ X ≤ 200) = P(X ≤ 200 ) - P(X ≤ 100 ) = .9375 - .7499 = .18761b. µ = = 72. 15 , σ = 72.15.01386P(X > µ + 2σ) = P(X > 72.15 + 2(72.15)) = P(X > 216.45) =1−−(216.45)(.01386)−2.9999[ 1−e ] = e = . 0498~c. .5 = P(X ≤ µ ~ − ( µ )(.01386)−(µ )(.01386)) ⇒ 1−e = .5 ⇒ e = . 5− µ ~ (.01386) = ln(. 5) = .693 ⇒ µ ~ = 50~161. Mean = = 25, 000λimplies λ = .00004(.00004)(20,000)a. P(X > 20,000) = 1 – P(X ≤ 20,000) = 1 – F(20,000; .00004) = e− = . 4491.2P(X ≤ 30,000) = F(30,000; .00004) = e− = . 699P(20,000 ≤ X ≤ 30,000) = .699 - .551 = .1481b. = = 25, 000λσ , so P(X > µ + 2σ) = P( x > 75,000) =1 – F(75,000;.00004) = .05.Similarly, P(X > µ + 3σ) = P( x > 100,000) = .018150