practice2+sol

# practice2+sol - Practice Problems for Exam 2 Problem 1 A...

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 4

practice2+sol - Practice Problems for Exam 2 Problem 1 A...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online