{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Test 5 Solution

# Test 5 Solution - Stat 230 Quiz#5 1 Let X be a continuous...

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

Stat 230 Quiz #5 November 24, 2004 1. Let X be a continuous random variable with probability density function (p.d.f.) f ( x ) = kx 2 for 0 < x < 2, otherwise f ( x ) = 0. (a) [ 3 marks ] Show that k = 3 / 8. Soln : 1 = R + -∞ f ( x ) dx = k R 2 0 x 2 dx = 8 k/ 3, so k = 3 / 8. (b) [ 3 marks ] Find the cumulative distribution function (c.d.f.) F ( x ) of X . Soln : F ( x ) = 0 if x 0 and F ( x ) = 1 if x 2; for x (0 , 2), F ( x ) = Z x 0 3 u 2 8 du = x 3 8 . (c) [ 3 marks ] Find P (1 < X < 3). Soln : P (1 < X < 3) = F (3) - F (1) = 1 - 1 / 8 = 7 / 8. (or P (1 < X < 3) = R 3 1 f ( x ) dx = R 2 1 (3 x 2 / 8) dx = 7 / 8) (d) [ 3 marks ] Find V ar ( X ). Soln : E ( X ) = R 2 0 x (3 x 2 / 8) dx = 3 × 2 4 / (4 × 8) = 3 / 2; E ( X 2 ) = R 2 0 x 2 (3 x 2 / 8) dx = 3 × 2 5 / (5 × 8) = 12 / 5. So V ar ( X ) = 12 / 5 - (3 / 2) 2 = 3 / 20 = 0 . 15. (e) [ 3 marks ] Let Y = ln( X ). Find the p.d.f. of Y . Soln : The range of Y is ( -∞ , ln(2)). For y ( , ln(2)), we have F Y ( y ) = P ( Y y ) = (ln( X ) y ) = P ( X exp( y )) = F X (exp( y )) = exp(3 y ) / 8 . So the p.d.f. of Y is given by f Y ( y ) = dF Y ( y ) /dy = 3 exp(3 y ) / 8 for y ( , ln(2)), otherwise f Y ( y ) = 0. 2. Suppose X follows a normal distribution N ( μ, σ 2 ) with μ = 1 and σ 2 = 4. Let F ( z ) be the cumulative distribution function (c.d.f.) of the standard normal distribution

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.

{[ snackBarMessage ]}

### Page1 / 2

Test 5 Solution - Stat 230 Quiz#5 1 Let X be a continuous...

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

View Full Document
Ask a homework question - tutors are online