QUIZ4 with solution

# QUIZ4 with solution - x âˆ âˆ âˆ âˆ âˆž âˆž-âˆž-âˆž = = = 3...

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Quiz 4 AMS310.01 Monday October 13 Print your name here_____________ ____1. Suppose f(x) = 3 for 0< X<1/3 and f(x) = 0 otherwise. Find the mean value of x. Hint = xdx 3 = 3x 2 /2. Suppose that Z has standard normal density (Z~N(0,1)). The p.d.f. of Z is f(z) = 2 / 2 2 1 x e - π for < < - x . Use Table 3 for the Standard Normal Distribution Function to answer Questions 2 through 4 below. _____ 2. P( Z< 1.28) = _____3. P( Z > 2.33) =____ _____ 4. z 0.1 = i.e., Find c such that P(Z> c) = 0.1.

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Quiz 4 SOLUTIONS __ 1/6__ 1. Suppose f(x) = 3 for 0< X<1/3 and f(x) = 0 otherwise. Find the mean value of
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Unformatted text preview: x. . âˆ« âˆ« âˆ« âˆ« âˆž âˆž-âˆž-âˆž + = + + = = 3 / 1 3 3 / 1 ) 3 ( ) ( xdx dx x xdx dx x xf Î¼ (3x 2 /2) = 3 (1/9)/ 2 = 1/6 _ 0.90____ 2. P(Z< 1.28) = Î¦ (1.28). From the Table 3: Î¦ (1.28)= 0.90 _ 0.01 ____3. P( Z > 2.33) =____ P(Z > 2.33) = 1- P(Z< 2.33) = 1- Î¦ (2.33) = 1- 0.99 = 0.01 (from Table 3: Î¦ (2.33)=0.99) __ 1.28 ___ 4. z 0.1 = i.e., Find c such that P(Z> c) = 0.1. P(Z > z 0.1 ) = 0.1. So P(Z< z 0.1 ) = Î¦ (z 0.1 )= 0.9 ( Z has standard normal c.d.f.) From Table 3(or from Quiz question 2): Î¦ ( 1.28 )= 0.90. So z 0.1 =1.28 2...
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