chapter 1

Chapter 4: Continuous Random Variables and Probability Distributions80.⎛ Y ⎞ 1 α 100 ⎛ Y ⎞a. E(Y) = 10 ⇒ E ⎜ ⎟ = = ; Var(Y) = ⇒ Var ⎜ ⎟ =⎝ 20 ⎠ 2 α + β 7 ⎝ 20 ⎠αβ⇒ α = 3, β = 3 , after some algebra.2α + β α + β + 1( ) ( )⎛ 12 ⎞ ⎛ 8 ⎞b. P(8 ≤ X ≤ 12) = F ⎜ ;3,3⎟− F⎜;3,3⎟ = F(.6;3,3) – F(.4; 3,3).⎝ 20 ⎠ ⎝ 20 ⎠The standard density function here is 30y 2 (1 – y) 2 ,∫.62so P(8 ≤ X ≤ 12) = 30 ( 1−y) dy = . 365.42y .c. We expect it to snap at 10, so P( Y < 8 or Y > 12) = 1 - P(8 ≤ X ≤ 12)= 1 - .365 = .665.1002800=128Section 4.681. The given probability plot is quite linear, and thus it is quite plausible that the tensiondistribution is normal.82. The z percentiles and observations are as follows:percentile observation-1.645 152.7-1.040 172.0-0.670 172.5-0.390 173.3-0.130 193.00.130 204.70.390 216.50.670 234.91.040 262.61.645 422.6lifetime400300200-2 -1 0 1 2z %ileThe accompanying plot is quite straight except for the point corresponding to the largestobservation. This observation is clearly much larger than what would be expected in a normalrandom sample. Because of this outlier, it would be inadvisable to analyze the data using anyinferential method that depended on assuming a normal population distribution.156