{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hw10sol

# hw10sol - Math 361/Stat 351 X1 Homework 10 Solutions Spring...

This preview shows pages 1–3. Sign up to view the full content.

Math 361/Stat 351 X1 Homework 10 Solutions Spring 2003 Graded problems: 1(d), 3(c), 4(b), 7 Problem 1. [4.2.4] Suppose component lifetimes are exponentially distributed with mean 10 hours. Find (a) the probability that a component survives 20 hours; (b) the median component lifetime (see p. 165 for a definition of the median of a distribution); (c) the SD of component lifetime; (d) the probability that the average lifetime of 100 independent components exceeds 11 hours. Solution. (a) We need to compute P ( X 20), where X is the life time of the component. We are given that X has exponential distribution with mean E ( X ) = 10 (measured in hours). Since for an exponential( λ ) distribution E ( X ) = 1 , we must have λ = 1 /E ( X ) = 1 / 10. Thus P ( X > 20) = 1 - P ( X < 20) = 1 - F (20) = 1 - (1 - e - (1 / 10) · 20 ) = e - 2 , using the formula F ( x ) = 1 - e - λx ( x > 0) for the c.d.f. of an exponential distribution. (b) The median lifetime is the value of t for which P ( X t ) = 1 / 2 = P ( X t ). Now, by the same argument as in (a), we have for t 0 P ( X t ) = e - 5 / 10 , and setting this value equal to 1 / 2 we get e - t/ 10 = 1 / 2 or t = 10 ln 2 = 6 . 93. (c) Since the lifetime X is exponentially distributed with parameter λ = 1 / 10, its standard deviation is, by the formula for the variance of an exponential distribution, SD ( X ) = p Var( X ) = p 1 2 = 1 = 10 . (d) Let X 1 , X 2 , . . . , X 100 be the lifetimes of the 100 independent components. The random variables X i are i.i.d., each having exponential distribution with λ = 1 / 10 and μ = E ( X 1 ) = 10, σ = SD ( X 1 ) = 10. We need to compute the probability P ( A 100 > 11), where A 100 = 1 100 100 X i =1 X i = 1 100 S 100 is the average lifetime (in the “naive” sense of average) of the 100 components. By normal approx- imation this is P ( A 100 > 11) = 1 - P ( A 100 11) = 1 - P ( S 100 1100) 1 - Φ 1100 - 100 · 10 100 · 10 = 1 - Φ(1) = 0 . 1587 . Problem 2. [4.R:4] Let X be a random variable with density f ( x ) = 0 . 5 e -| x | ( -∞ < x < ). Find (a) P ( X < 1), (b) E ( X ) and SD ( X ), (c) the c.d.f. of X 2 . Solution. P ( X < 1) = Z 1 -∞ f ( x ) dx = 1 2 Z 1 -∞ e -| x | dx = 1 2 Z 0 -∞ e x dx + 1 2 Z 1 0 e - x dx (a) = 1 2 e x 0 -∞ + 1 2( - 1) e - x 1 0 = 1 2 + 1 2 ( - e - 1 + 1 ) = 1 - 1 2 e . 1

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

View Full Document
E ( X ) = Z -∞ xf ( x ) dx = Z -∞ x 1 2 e -| x | dx = 0 (since integrand is an odd function) (b) E ( X 2 ) = Z -∞ x 2 f ( x ) dx = Z -∞ x 2 1 2 e -| x | dx = Z 0 x 2 e - x dx (by symmetry) = 2 (from integral table or by integration by parts) Thus, SD ( X ) = p E ( X 2 ) - E ( X ) 2 = 2 - 0 2 = 2.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 5

hw10sol - Math 361/Stat 351 X1 Homework 10 Solutions Spring...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online