chapter 1

Chapter 3: Discrete Random Variables and Probability Distributions92. p(y) = P(Y = y) = P(y trials to achieve r S’s) = P(y-r F’s before r th S)= nb(y – r;r,p) =⎛ y − 1⎞r⎜ ⎟ p (1 − p)⎝ r − 1⎠y−r, y = r, r+1, r+2, …93.a. b(x;15,.75)b. P(X > 10) = 1 - B(9;15, .75) = 1 - .148c. B(10;15, .75) - B(5;15, .75) = .314 - .001 = .313d. µ = (15)(.75) = 11.75, σ 2 = (15)(.75)(.25) = 2.81e. Requests can all be met if and only if X ≤ 10, and 15 – X ≤ 8, i.e. if 7 ≤ X ≤ 10, so P(allrequests met) = B(10; 15,.75) - B(6; 15,.75) = .31094. P( 6-v light works) = P(at least one 6-v battery works) = 1 – P(neither works)= 1 –(1 – p) 2 . P(D light works) = P(at least 2 d batteries work) = 1 – P(at most 1 D batteryworks) = 1 – [(1 – p) 4 + 4(1 – p) 3 ]. The 6-v should be taken if 1 –(1 – p) 2 ≥ 1 – [(1 –p) 4 + 4(1 – p) 3 ].Simplifying, 1 ≤ (1 – p) 2 + 4p(1- p) ⇒ 0 ≤ 2p – 3p 3 ⇒ p ≤ 3 2 .95. Let X ~ Bin(5, .9). Then P(X ≥ 3) = 1 – P(X ≤ 2) = 1 – B(2;5,.9) = .99196.a. P(X ≥ 5) = 1 - B(4;25,.05) = .007b. P(X ≥ 5) = 1 - B(4;25,.10) = .098c. P(X ≥ 5) = 1 - B(4;25,.20) = .579d. All would decrease, which is bad if the % defective is large and good if the % is small.97.a. N = 500, p = .005, so np = 2.5 and b(x; 500, .005) =&p(x; 2.5), a Poisson p.m.f.b. P(X = 5) = p(5; 2.5) - p(4; 2.5) = .9580 - .8912 = .0668c. P(X ≥ 5) = 1 – p(4;2.5) = 1 - .8912 = .1088123