chapter 1

Chapter 5: Joint Probability Distributions and Random Samples71.121 1 2 21 1 2 2W0a. M a X + a X + W xdx = a X + a X + 72 ,= ∫soE(M) = (5)(2) + (10)(4) + (72)(1.5) = 158m2M2 2 2 2 2 2( 5) (.5) + ( 10) ( 1) + ( 72) (.25) = 430. 25σ =, σ = 20. 74M⎛ 200 −158⎞b. P ( M ≤ 200) = P⎜Z≤ ⎟ = P(Z ≤ 2.03) = . 9788⎝ 20.74 ⎠72. The total elapsed time between leaving and returning is T o = X 1 + X 2 + X 3 + X 4 , withE ( ) = 40, σ 2 = 40 , σ = 5. 477T oT oT o. T o is normally distributed, and the desired value tis the 99 th percentile of the lapsed time distribution added to 10 A.M.: 10:00 +[40+(5.477)(2.33)] = 10:52.7673.a. Both approximately normal by the C.L.T.b. The difference of two r.v.’s is just a special linear combination, and a linear combinationX − Y has approximately a normal2 22 8 6µ = and σ = + = 2.629, σ = 1. 621X −YX −YX−Y40 35of normal r.v’s has a normal distribution, sodistribution with 5− 1−5 1−5P −1 ≤ X − Y ≤ 1 ≈&P⎜≤ Z ≤⎝1.62131.6213= P ( −3.70≤ Z ≤ −2.47)≈ .0068⎛⎞c. ( ) ⎟⎠⎛ 10 − 5 ⎞P & ⎜ ⎟This probability is⎝ 1.6213⎠− µd. ( X − Y ≥ 10 ) ≈ P Z ≥ = P(Z ≥ 3.08) = .0010.quite small, so such an occurrence is unlikely if µ1 2= 5 , and we would thus doubtthis claim.74. X is approximately normal with = (50)(.7) 352µ and σ = (50)(.7)(.3) 10. 51=is Y with µ2= 30 and σ 2= 212 . Thus = 5 X −Y⎛ −100 ⎞p − ≤ X − Y ≤ 5 ≈ P⎜≤ Z ≤ ⎟ =⎝ 4.74 4.74 ⎠1=2X −Y= 22. , soµ and σ 5( 5 ) P(−2.11≤ Z ≤ 0) = . 4826, as198