This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Old Midterm Answers First Midterm 1) Assumptions: Something to the effect that exam times are like a sam ple without replacement from all possible exam times. Thus all parenleftBig 18 4 parenrightBig sets of 4 exams are equally likely, and we need to count how many ways there are of having a schedule with all 4 exams on different days. parenleftBig 6 4 parenrightBig is the number of ways to choose 4 distinct days out of the 6 days, and each day there are 3 possible exam times. This is like choosing a rank and then a suit from draws from a deck of cards. The answer is thus: ( 6 4 ) 3 4 ( 18 4 ) An other solution: Find the probability that the successive exams don’t fall on a day already chosen. So once the first exam is set, the chance the second exam is on a different day is 15 17 , and the chance the third is on a different day than the first two is 12 16 , so the answer is: 15 17 × 12 16 × 9 15 2) a) i) parenleftBig 10 3 parenrightBig 1 2 3 1 2 7 from the binomial formula. ii) binomial(10, 1 2 ) b) i) μ = np = 50, σ = √ npq = 5. P( X ≤ 55)= Φ( 55+ . 5 50 5 ) = Φ(1 . 1) ii) Normal(50,5 2 ) (approx of binomial(100, 1 2 ) c) i) Two ways of solving this: I) P ( Y =3 and X =50) P ( X =50) The numerator becomes P( Y = 3) × P(47 black balls in 90 draws), and after much cancelling the answer is ( 10 3 )( 90 47 ) ( 100 50 ) which is suggestive of the hypergeometric distribution. II) If exactly 50 black balls were drawn out of the 100, we can look at the first 10 as a sample without replacement from these 100 draws, and we get ( 50 3 )( 50 7 ) ( 100 10 ) Cancelling will show that these answers are the same....
View
Full Document
 Spring '08
 MCCULLOUGH
 Normal Distribution, Binomial distribution, #, 6 days, geom, 17 15 9 day

Click to edit the document details