{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

final(spring2002)--solutions

final(spring2002)--solutions - 0 Problem 1.A random sample...

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

Problem 1.A random sample of size n = 100 was drawn from a population. A histogram of the data is shown in the figure. a. (5 pts) Multiple choice. The histogram indicates the distribution is: A. normal B. chi-squared C. light-tailed D. heavy-tailed b. (5 pts) Multiple choice. Let X be a random variable for an observation from the population discussed above. Suppose the population mean μ and population variance σ 2 are known. After assuming that the population is normal a probablist calculates - = σ μ 5 ) 5 ( Z P X P . Assuming she makes no calculation errors, she: A. overestimates the true probability B. underestimates the true probability C. gets the true probability exactly right D. still cannot calculate the true probability because of random variation. Problem 2. Suppose that among engineering students, 13% of men and 9% of women are left-handed. a. (5 pts) Assuming 22% of the engineering students are women, what is the probability that a randomly selected engineering student is left-handed? Suppose an engineering student is selected at random. Let L be the event that the student is left-handed and W be the event that the student is a woman. Then: % 12 . 12 78 . 0 13 . 0 22 . 0 09 . 0 ) ( ) | ( ) ( ) | ( ) ( = + = + = W P W L P W P W L P L P b. (5 pts) Suppose in the course of her Cornell undergraduate career, a woman has 5 boyfriends who are all engineering students. If the boyfriends are randomly selected, what is the probability that there would be 3 or more left-handers in her sample of 5 boyfriends? Let X be a random variable for the number of lefties in a random selection of 5 engineering boyfriends. Then, assuming X ~ Binomial(5, 0.13), we can calculate that probability as: 0179 . 0 87 . 0 13 . 0 5 5 87 . 0 13 . 0 4 5 87 . 0 13 . 0 3 5 ) 3 ( 0 5 1 4 2 3 = + + = X P Problem 3. (5 pts) The probability density function for a random variable Y is f ( y ) = 0.1*( y + 3) for 1 < y < 3 and f ( y ) = 0 otherwise. Find F (3). F (3) = 1 [Obvious.] Or: 1 ) 3 2 / 1 ( 9 2 3 1 . 0 3 2 1 . 0 ) 3 ( 1 . 0 ) ( ) 3 ( 2 3 1 2 3 1 3 = + - + = + = + = = - y y dy y dy y f F 1 -10 -5 0 5 10 0.0 0.10 0.20 x Relative Frequency

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

View Full Document
Problem 4. (40 points through problem 4) A newspaper reports that on the night of June 7, the glorious Perseid meteor shower can be viewed. It is claimed that an observer should be able to view an average of 20 meteors per hour. Assume that meteor sightings are independent amd the rate is constant throughout the night. a. (5 pts) Let X be a random variable for the amount of time an observer would have to wait until first observing a meteor. Suppose the newspapers' claim that the rate is 20 sighting /hr is correct. Then, find the 95th percentile of X . Assume X ~ Exponential(20). Then, the 95 th percentile would be the number η , such that 95 . 0 20 0 20 = - η dx e x .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 7

final(spring2002)--solutions - 0 Problem 1.A random sample...

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

View Full Document
Ask a homework question - tutors are online