Disclaimer due to some extremely pressing deadlines i

• Notes
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**********************DISCLAIMER******************** Due to some extremely pressing deadlines I have (it is now 4am), I gratefully accepted an offer of someone who agreed to write up these solutions. I just went through to check them and I made some mi- nor modifications to the solutions above this disclaimer, which mainly

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consisted of adding some details in order to clarify some things. How- ever, I have run out of time. I’m going to post these “as is” for people who might want to check their answers on Wednesday morning be- fore class. I hope to get back to checking the rest of the problems (below) after class on Wednesday. 7. n = 10, x = 230, s = 15 Because of the small sample size, we need to assume normality in order to do this problem. The critical value comes from the t -curve with n - 1 = 9 degrees of freedom. X ± t α/ 2 s n 230 ± 1 . 383 15 10 (223 . 44 , 236 . 56) 8. a. Since X and Y are independent, their joint pdf is the product of th e two marginal pdfs: f ( x, y ) = f X ( x ) · f Y ( y ) = e - x · e - y = e - ( x + y ) , x > 0 , y > 0 and zero otherwise. b. You could do the double integral of f ( x, y ) as x and y both go from 0 to 1 which is equivalent to multiplying the two single integrals in this case because of independence. P ( X 1) = Z 1 0 e - x dx = 1 - e - 1 So P ( X 1 , Y 1) indep = P ( X 1) · P ( Y 1) = (1 - e - 1 ) · (1 - e - 1 ) 0 . 39958 c. The covariance is zero since they are independent!
9. n = 1525, x = 191 . 7, s = 41 . 0 X ± z α/ 2 s n 191 . 7 ± (2 . 17) 41 1525 (189 . 422 , 193 . 978) 10. The sample mean is 20. The sample variance is 5.2. Because of the sma ll sample, we will have to assume normality. (A probability plot would be nice if this were a hw problem, but you wouldn’t have to do this on an exam.) Because of the small sample and the fact that we have s instead of σ , we use a t -critical value based on n - 1 = 5 degrees of freedom.

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