finalans

# finalans - Answers to Final Exam Review Questions Question...

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

Answers to Final Exam Review Questions Question 1 : This problem is a Bayes Law problem. The forward tree branches first on “will” and “won’t” date George; the second branch is “is free” and “isn’t free”. When you reverse the tree you find that the () P will date George says"free" = + = . .. . 075 075 225 25 . Question 2 : Here we have X = 13, Median = 14, s = 592 . ; the 98% C.I. is 729 1871 ., . A normal quantile plot, sometimes called a normal probability plot, is the technique I am looking for. This is the one that creates a straight line if the data is normally distributed. Question 3 : Part (a) is hypergeometric, with N r n x or == = = 10 6 3 2 3 ,,, . The answer is two thirds. Part (b) is poisson with µ = 33 5 / and x = 1. The answer is .0787. Question 4 : This is a straightforward normal probability calculation. The answer for part (a) is z = =− 0 5 31 16 . . . and Pz >− = 16 5636 . The answer to part (b) is −= z X . . . . 35 39 305 31 , or X = 293 Question 5 : In part (a) the mean is xf x 25 and the variance is ( ) σµ 2 2 105 = xf x . In part (b) you use the formula for the mean and variance of a linear transformation of a random variable; that is Y ab X X =+ ≡ .10 . Therefore the mean of the phone bill is \$2.50, and the variance is 1.05. In part (c) we have a problem involving the distribution of the mean. The question asks the probability that

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

View Full Document
() Px P x n x i i >= = 72 72 30 240 . . We know that Ex == µ 250 . and that σ x n = 105 30 187 . . . Therefore the z value is z = =− 240 250 187 53 .. . . and the answer is Pz >− = 53 7019 . In part (d) the answer is yes; because the sample size is large (n=30) the sample means have a normal distribution even though the original x’s obviously did not. We needed the normal distribution to do our probability calculation. Question 6 Part (a) is a binomial with n = 5 and p = 18/38. Being a net winner means winning 3, 4, or 5 of your 5 bets. The probability of this is 5 3 474 526 5 4 474 526 5 5 474 526 4514 32 41 50 + + = . For part (b) you are asked to do a normal approximation to a binomial with n=25 and p = 18/38 = .474. The distribution of the number of successes in 25 trials is approximately normal with () ( ) 25 .474 11.84 EX n p = and 1 25 .474 .526 2.497 X np p = × × = Being a net winner means winning 13 or more of your bets. The z value is 12.5 11.84 .26 2.497 z . The answer is ( ) .26 .5000 .1026 .3974 > =−= .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

finalans - Answers to Final Exam Review Questions Question...

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

View Full Document
Ask a homework question - tutors are online