chapter 1

Chapter 5: Joint Probability Distributions and Random Samples53. µ = 50, σ = 1.2a. n = 9⎛ 51−50 ⎞P( X ≥ 51) = P ⎜Z≥ ⎟ = P(Z ≥ 2.5) = 1−.9938= . 0062⎝ 1.2 / 9 ⎠b. n = 40⎛ 51−50 ⎞P( X ≥ 51) = P ⎜Z≥ ⎟ = P(Z ≥ 5.27) ≈ 0⎝ 1.2 / 40 ⎠54.σx.85µ X, σ x = = = . 17n 5a. = µ = 2. 65⎛ 3.00 − 2.65 ⎞P( X ≤ 3.00)= P ⎜Z≤ ⎟ = P(Z ≤ 2.06) = . 9803⎝ .17 ⎠P(2.65 ≤ X ≤ 3.00)= = P ( X ≤ 3.00) − P(X ≤ 2.65) = . 4803⎛ 3.00 − 2.65⎞.35b. P( X ≤ 3.00)= P ⎜Z≤ ⎟ = . 99 implies that = 2.33,⎝ .85/ n ⎠85/ nwhich n = 32.02. Thus n = 33 will suffice.fromµ σ = npq = 3. 46455. = np = 20a.⎛ 24.5 − 20 ⎞P( 25 ≤ X ) ≈ P ⎜ ≤ Z ⎟ = P(1.30≤ Z ) = . 0968⎝ 3.464 ⎠b.⎛14.5− 20 25.5 − 20 ⎞P( 15 ≤ X ≤ 25) ≈ P ⎜ ≤ Z ≤ ⎟⎝ 3.464 3.464 ⎠= P ( −1.59≤ Z ≤ 1.59) = .888256.a. With Y = # of tickets, Y has approximately a normal distribution with µ = λ = 50 ,⎛ 34.5 − 50 70.5 − 50 ⎞σ = λ = 7.071, so P( 35 ≤ Y ≤ 70) ≈ ⎜ ≤ Z ≤ ⎟⎝ 7.071 7.071 ⎠≤ Z ≤ 2.90) = .9838P = P( -2.192b. Here µ = 250 , σ = 250, σ = 15. 811⎛ 224.5 − 250 275.5 − 250 ⎞⎜≤ Z ≤⎟⎝ 15.811 15.811 ⎠, so P( 225 ≤ Y ≤ 275) ≈P = P( -1.61 ≤ Z ≤ 1.61) = .8926193