Random Variables - Solution

Random Variables - Solution - (b) ? , ? = 12 2 7 + 6 7 2...

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CE93 Engineering Data Analysis Random Variables Practice Problems (All problems taken from the textbook “Probability and Statistics for engineers and Scientists, Fourth Edition” by Sheldon M. Ross): Problem 4.1. (PMF’s, Combinatorics) P 1 = 5/10 P 2 = 5/10 × 5/9 = .2778 P 3 = 5/10 × 4/9 × 5/8 = .1389. P 4 = 5/10 × 4/9 × 3/8 × 5/7 = .0595 P 5 = 5/10 × 4/9 × 3/8 × 2/7 × 5/6 = .0198, P 6 = 5/10 × 4/9 × 3/8 × 2/7 × 1/6 = .0040, P 7 = P 8 = P 9 = P 10 = 0 where P i = P ( X = i ) . Problem 4.4 (CDF’s, Probabilities from Information about CDF’s) a) b) 1 − F ( 1/2 ) = ¾ c) F ( 4 ) F ( 2 ) = 1/12 d) lim ℎ→ 0 ? (3 ) = 11/12 e) F ( 1 ) lim ℎ→ 0 ? (1 ) = 2/3 − 1/2 = 1/6 Problem 4.6 (PDF’s) Note first that since f ( x ) dx = 1, it follows that λ = 1/100; therefore, ?±²³?² 150 50 = ´ 1 2 − ´ 3 2 = 0.3834 Also, ?±²³?² = 1 − ´ 1 = 0.6321 100 0 . Problem 4.10 (PDF’s – part c is challenging but is still reasonable. For a hint, see example 4.3.C) (a) Show that the double integral of the joint density equals 1.
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Unformatted text preview: (b) ? , ? = 12 2 7 + 6 7 2 (c) ? , ?? = 6 3 7 + 3 3 14 1 1 ? = 15 56 Problem 4.13 (PDFs) (i) 2 1 ? = 2 , 0 < < 1 (ii) 2 ? = 2 , 0 < < 1 (iii) No, since the product of the individual densities is not equal to the joint density. Problem 4.19 (Conditional PDFs) (a) ? = ? , , ? = 2 + 3 14 1 , 0 < < 2. Hence, ? ? | ? = 12 2 + 6 4 + 3 , 0 < < 1. Problem 4.32 (Expected Value) (a) E(4 + 16X + 16X 2 ) = 164 (b) E(X 2 + X 2 + 2X + 1) = 21 Problem 4.45 (Marginal PDFs, Expected Value, Variance and Covariance) 1 = 3 16 = 0 1 8 = 1 5 16 = 2 3 8 = 3 0 2 = 1 2 = 1 1 2 = 2 0 ? 1 = 30 16 ? 1 = 19 4 15 8 2 = 1.234 ? 2 = 3 2 ? 2 = 0.25...
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Random Variables - Solution - (b) ? , ? = 12 2 7 + 6 7 2...

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