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

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AMS 310.01 Fall 2003 Homework 4 Solutions 4.30 (a) The mean µ is given by 0(0.4) 1(0.3) 2(0.2) 3(0.1) 1 μ = + + + = (b) The variance 2 σ is given by 2 2 2 2 2 (0 1) (0.4) (1 1) (0.3) (2 1) (0.2) (3 1) (0.1) 1 σ = - + - + - + - = 4.34 Using the formula, 0 1 ( 1) 2 n i i n n = = + and 2 0 1 ( 1)(2 1) 6 n i i n n n = = + + We find 0 1 1 2 n i n i n μ = + = = , and 2 2 2 2 2 1 1 ( 1) ( 1)( 1) ( 1) ' ( 1)(2 1) [ ] 6 2 12 12 n n n n n n n n σ μ μ + + - - = - = + + - = = 4.36 (a) The mean µ is given by 5 . 2 32 1 5 32 5 4 32 10 3 32 10 2 32 5 1 32 1 0 = + + + + + = μ (b) 5 . 2 ) 5 (. 5 = = = p n μ 4.40 (a) The probabilities are : 4 4 0 3 4 ( 0) 8 56 3 P X ���� ���� ���� = = = �� �� �� , 4 4 1 2 24 ( 1) 8 56 3 P X ���� ���� ���� = = = �� �� �� 4 4 2 1 24 ( 2) 8 56 3 P X ���� ���� ���� = = = �� �� �� , 4 4 3 0 4 ( 3) 8 56 3 P X ���� ���� ���� = = = �� �� �� Thus, 4 24 24 4 0 1 2 3 1.5 56 56 56 56 μ = + + + = 2 2 2 2 2 4 24 24 4 30 (0 1.5) (1 1.5) (2 1.5) (3 1.5) 0.5357 56 56 56 56 56 σ = - + - + - + - = = (b) From the Special formulas, we have 4 3 1.5 8 a n N μ = = = , 2 2 2 ( )( ) 3 4 4 5 240 0.5357 ( 1) 8 7 448 n a N a N n N N σ - - ��� = = = = -
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4.42 The tail probabilities and upper bounds are Number of sd’s Upper Bound of tail probabilities Tail probabilities from binomial (16; 1/2) 1 1 0.4544 2 1/4 0.0768 3 1/9 0.0042 The upper bound of the tail probability comes from 2 1 (| | ) P X k k μ σ - where k is the number of standard deviations. An example of the calculation of the tail probabilities for the binomial distribution with
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