hw4sol_fall03 - AMS 310.01 Fall 2003 4.30 Homework 4...

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Unformatted text preview: AMS 310.01 Fall 2003 4.30 Homework 4 Solutions (a) The mean is given by = 0(0.4) + 1(0.3) + 2(0.2) + 3(0.1) = 1 (b) The variance 2 is given by 2 = (0 - 1) 2 (0.4) + (1 - 1) 2 (0.3) + (2 - 1)2 (0.2) + (3 - 1)2 (0.1) = 1 4.34 Using the formula, n 1 1 n(n + 1) i 2 = n(n + 1)(2n + 1) and 2 6 i =0 i =0 n 1 n +1 i= We find = , and n i=0 2 11 (n + 1) 2 (n + 1)(n - 1) (n 2 - 1) 2 2 = 2 '- = n(n + 1)(2n + 1) - [ ] = = n6 2 12 12 n i= 4.36 (a) The mean is given by = 0 (b) 1 5 10 10 5 1 + 1 + 2 + 3 + 4 + 5 = 2.5 32 32 32 32 32 32 = n p = 5(.5) = 2.5 4.40 (a) The probabilities are : 4 4 0 3 = 4 P ( X = 0) = 8 56 3 4 4 1 2 24 P ( X = 2) = = 8 56 3 Thus, , 4 4 1 2 = 24 P ( X = 1) = 8 56 3 4 4 0 3 4 P ( X = 3) = = 8 56 3 , 4 24 24 4 = 0 + 1 + 2 + 3 = 1.5 56 56 56 56 4 24 24 4 30 2 = (0 - 1.5) 2 + (1 - 1.5) 2 + (2 - 1.5) 2 + (3 - 1.5) 2 = = 0.5357 56 56 56 56 56 (b) From the Special formulas, we have a 4 = n = 3 = 1.5 , N 8 2 = n ( N - a )( N - n) 3 240 a 4 4 5 = = = 0.5357 2 N ( N - 1) 82 7 448 4.42 The tail probabilities and upper bounds are Number of sd's Upper Bound of tail Tail probabilities from binomial (16; 1/2) 0.4544 0.0768 0.0042 probabilities 1 1 2 1/4 3 1/9 The upper bound of the tail probability comes from P (| X - | k ) where k is the number of standard deviations. 1 k2 An example of the calculation of the tail probabilities for the binomial distribution with n=16 and p=1/2 follows for the case k=2, = np = 8 Thus, we need , 2 = np(1 - p ) = 4 , =2 P ( X P 4) + P ( X 12) when X has distribution b(x;16,1/2). From the Table 1, this is 0.0384 + 1 - 0.9616 = 0.0768 4.50 (a) F(4;7) = 0.173 (b) f(4;7) = F(4;7) - F(3;7) = (0.173) - (0.082)=0.091 19 (c) k =6 f (k ;8) = F (19;8) - F (5;8) = 1.0 - (0.191) = 0.809 4.54 n = 400 , p = 0.008 , np = 3.2 f(4;3.2) = F(4;3.2) - F(3;3.2) = (0.781) - ( 0.603) =0.178 4.64 The Poisson process with = 2 applies. (a) We use Table 2 with = 2(2) = 4. f (5;4) = F (5;4) - F (4;4) = 0.785 - 0.629 = 0.156 (b) 1 - F (7;4) = 1 - 0.949 = 0.051 (c) For the separate one hour periods, the probabilities are f ( 2;2) = 2 2 -2 e = .2707 2! f (3;2) = 2 3 -2 e = .1804 3! The intervals do not overlap so the counts are independent and we multiply the two probabilities f ( 2;2) f (3;2) = (.2707) (.1804) = .0488 ...
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This homework help was uploaded on 01/31/2008 for the course AMS 310.01 taught by Professor Mendell during the Fall '03 term at SUNY Stony Brook.

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