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Unformatted text preview: Practice Problems for Exam 2 Problem 1. A amount X in mg of a certain chemical in a certain product is a random variable with density given by f ( x ) = ( c (2 x )( x 1) 2 1 ≤ x ≤ 2 otherwise . Find the constant c , the probability that the quantity is between 1 . 2 and 1 . 8 . Find also the mean and the variance of X . Solution. Since the integral of f on [1 , 2] must be 1, it follows that c = 12. Then P (1 . 2 ≤ X ≤ 1 . 8) = R 1 . 8 1 . 2 f ( x ) dx = 0 . 792. The mean is 8 / 5 and the variance is 1 / 25. Problem 2. The radius X and the height Y in mm of a cylinder in a certain car are independent random variables with densities given by f X ( x ) = ( 64 x ( x 100)(101 x ) / 535 100 ≤ x ≤ 100 . 5 otherwise f Y ( y ) = ( 192 y 2 (22 y ) / 68813 20 ≤ y ≤ 20 . 5 otherwise . Find the mean value and the standard deviation of the area of the base and volume of the cylinder. Solution. The mean of the area A = πX 2 is given by μ A = μ πX 2 = Z 100 . 5 100 πx 2 f X ( x ) dx = 31612 . 7 For the variance we have σ 2 A = μ π 2 X 4 ( μ πX 2 ) 2 and μ A 2 = μ π 2 X 4 = Z 100 . 5 100 π 2 x 4 f X ( x ) dx = 9 . 9937 × 10 8 . and σ 2 A = 5891 . 35. Since X and Y are independent, the mean of the volume of the cylinder is μ V = μ AY = πμ A μ Y = 639846 and the variance σ 2 V = μ A 2 μ Y 2 ( μ A μ Y ) 2 = 2 . 31178 × 10 7 ....
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 Spring '08
 CITRIN
 Probability, Standard Deviation, Electromagnet, Probability theory, density function

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