chapter 1

Chapter 4: Continuous Random Variables and Probability Distributions100.a. F(X) = 0 for x < 1 and = 1 for x > 3. For 1 ≤ x ≤ 3, F ( x)= ∫ −∞f ( y)dy⎛ ⎞1+ x 3 11∫ 0dy⋅ dy = 1.51⎜1−−∞⎟1 22 y ⎝ x ⎠b. P(X ≤ 2.5) = F(2.5) = 1.5(1 - .4) = .9; P(1.5 ≤ x ≤ 2.5) =F(2.5) – F(1.5) = .43 3 1 3 3 13c. E(X) = = ∫ x ⋅ ⋅ dx = = 1.5ln( )] 1= 1. 6481 22 2∫ dx xx1xd. E(X 2 32 3 1 3 3) = = ∫ x ⋅ ⋅ dx = = 31 22 2∫ dx , so V(X) = E(X 2 ) – [E(X)] 2 = .284,x1σ =.553⎧ 0⎪e. h(x) = ⎨x−1.5⎪⎩ 131≤x ≤1.51.5 ≤ x ≤ 2.52.5 ≤ x ≤ 32.53so E[h(X)] = = ∫ ( x −1.5) ⋅ ⋅ dx + ∫ 1⋅⋅ dx . 26721x=1.52 2.52321x101.a.0.40.3f (x)0.20.10.0-2-10x123b. F(x) = 0 for x < -1 or == 1 for x > 2. For –1 ≤ x ≤ 2,3x 1 2 1 ⎛ x ⎞ 11F(x)= ∫ ( 4 − y ) dy = 4 +199⎜ x −3⎟−⎝ ⎠ 27c. The median is 0 iff F(0) = .5. Since F(0) = 11 27, this is not the case. Because 11 27< .5, themedian must be greater than 0.5d. Y is a binomial r.v. with n = 10 and p = P(X > 1) = 1 – F(1) = 27166