chapter 1

Chapter 4: Continuous Random Variables and Probability Distributions6.a.2f(x)10012x345412 4[ k[1− u ] du = ⇒ k =−132b. 1 =∫ k 1 − ( x − 3) ] dx = ∫234∫4c. P(X > 3) = 3 2[1 − ( x − 3) ] dx = . 5 by symmetry of the p.d.f3413/ 41332 32d. P ( 11 ≤ X ≤ ) = [1 − ( − 3) ] = [1 − ( ) ] = ≈ . 36744 ∫ x dx44 ∫ u du11/ 4e. P( |X-3| > .5) = 1 – P( |X-3| ≤ .5) = 1 – P( 2.5 ≤ X ≤ 3.5).5∫ − .541/ 4−1/447128= 1 - 3 2[1 − ( u ) ] du = ≈ . 3135167.a. f(x) = 101for 25 ≤ x ≤ 35 and = 0 otherwise35∫ dxb. P(X > 33) = . 2∫3533 10 1=2x ⎤⎥20 ⎦1c. E(X) = x ⋅ dx = = 302510352530 ± 2 is from 28 to 32 minutes:321 32=28 10 10 28∫1P(28 < X < 32) = dx = x] . 42∫ a+a101d. P( a ≤ x ≤ a+2) = dx = . 2 , since the interval has length 2.131