chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions50. X ~ Bin(10, .60)a. P(X ≥ 6) = 1 – P(X ≤ 5) = 1 - B(5;20,.60) = 1 - .367 = .633b. E(X) = np = (10)(.6) = 6; V(X) = np(1 – p) = (10)(.6)(.4) = 2.4;σ x = 1.55E(X) ± σ x = ( 4.45, 7.55 ).We desire P( 5 ≤ X ≤ 7) = P(X ≤ 7) – P(X ≤ 4) = .833 - .166 = .667c. P( 3 ≤ X ≤ 7) = P(X ≤ 7) – P(X ≤ 2) = .833 - .012 = .82151. Let S represent a telephone that is submitted for service while under warranty and must bereplaced. Then p = P(S) = P(replaced | submitted)⋅P(submitted) = (.40)(.20) = .08. Thus X,the number among the company’s 10 phones that must be replaced, has a binomial⎛10⎞8⎝ 2 ⎠2distribution with n = 10, p = .08, so p(2) = P(X=2) = ⎜ ⎟(.08)(.92) = . 147852. X ∼ Bin (25, .02)a. P(X=1) = 25(.02)(.98) 24 = .308b. P(X=1) = 1 – P(X=0) = 1 – (.98) 25 = 1 - .603 = .397c. P(X=2) = 1 – P(X=1) = 1 – [.308 + .397]x ; σ = npq = 25 (.02)(.98) = .49 = . 7d. = 25 (.02) = . 5x + 2 σ = .5 + 1.4 = 1.9 So P(0 = X = 1.9 = P(X=1) = .705.5(4.5) + 24.5(3)e. = 3. 03 hours2553. X = the number of flashlights that work.Let event B = {battery has acceptable voltage}.Then P(flashlight works) = P(both batteries work) = P(B)P(B) = (.9)(.9) = .81 We mustassume that the batteries’ voltage levels are independent.X∼ Bin (10, .81). P(X=9) = P(X=9) + P(X=10)1010⎛ ⎞⎜ ⎟⎝ 9 ⎠⎛10⎞⎝10⎠9(.81) (.19) + ⎜ ⎟( .81) = .285 + .122 = . 407109