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Unformatted text preview: STATISTICS 321 – Dr. Soma Roy Exam 2 review: Practice problems: SOLUTIONS 1. X =number of patients who recover, n = 4 and p = . 4. X ∼ binomial (4 ,. 4) (a) P ( X = 2) = 4 2 ! ( . 4) 2 ( . 6) 2 = . 3456 (b) P ( X ≥ 1) = 1 P ( X = 0) = 4 ! ( . 4) ( . 6) 4 = 1 . 1296 = 0 . 8704 (c) P ( X = 0) = . 1296 (d) E ( X ) = np = (4)( . 4) = 1 . 6 2. • P ( X < 11) = P ( X ≤ 10) = . 902 • P ( X > 22) = 1 P ( X ≤ 22) ≈ 1 1 = 0 3. Find E ( X ) E ( X ) = Z 5 1 1 18 x ( x + 1) dx = 1 18 " x 3 3 + x 2 2 # 5 1 = 3 Expected profits are $3000.00. 4. E ( X ) = Z 1 x 2 dx + Z 2 1 x (2 x ) dx = 1 E ( X 2 ) = Z 1 x 3 dx + Z 2 1 x 2 (2 x ) dx = 7 6 V ar ( X ) = E ( X 2 ) [ E ( X )] 2 = 7 6 1 = 1 6 E (2 X + 3) = 2 E ( X ) + 3 = 5 V ar (2 X + 3) = 2 2 V ar ( X ) = 4 V ar ( X ) = 2 3 5. ... (a) R ∞ ce x 3 dx = 1 ⇒ c = 1 3 (b) Find E(X) E ( X ) = Z ∞ x 1 3 e x 3 dx = 3( integration by parts ) To find P ( X > 5) P ( X > 5) = 1 P ( X < 5) = Z 5 1 3 e x 3 dx = e 5 3 (c) E ( X +2) =...
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This note was uploaded on 06/15/2010 for the course CHEM 124 taught by Professor Hascall during the Winter '08 term at Cal Poly.
 Winter '08
 Hascall

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