Homework4 - 3) = . 31+ . 40 = . 71 , P ( X 3) =

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Homework 4 Stat 3022 February 21, 2010 Problem 5.13 1. C is distributed Bin(10,.7), M is distributed Bin(10,.3) 2. > pbinom(3,10,.3,lower.tail=F) [1] 0.3503893 Problem 5.14 1. X the number of acution site visitors is, Bin(15,.5). 2. Symmetry tells us which is confirmed by R P ( X 8) =.5 pbinom(7,15,.5,lower.tail=F) [1] 0.5 Problem 5.15 Values from Problem 5.13 1. C has the mean (10)(.7)=7. M has the mean (.3)(10)=3. 2. Both M and C have the same standard deviation sqrt 10( . 7)( . 3) = 1 . 4491 3. With p=.9 follow the same formula as in b the standard deviation is .9487. With p=.99 the standard deviation is .3146. Thus we can conclude that as p approaches 1 the standard deviation approaches 0. Problem 5.16 1. The mean of X is (15)(.05)=7.5. The mean of ˆ p is .05. 1
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2. The mean of X increases with n. It is 75 with n=150, and 750 with n=1500. The mean of ˆ p is .5 for any value of n. Problem 5.32 SEE BRAD FOR SOLUTION 5.50 1. μ X =(4)(.31)+(3)(.4)+(2)(.2)+(1)(.04)=2.88, σ X = 1 . 1056 =1.0515 2. μ ¯ X =2.88, σ ¯ X = σ X 50 =.1487 3. P ( X
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Unformatted text preview: 3) = . 31+ . 40 = . 71 , P ( X 3) = pnorm(3,2.88,.1487,lower.tail=F)=.2098 Problem 5.52 1. X =(500)(.001)=.50, X = 249 . 75 =15.8 2. In the long run Joe will make about 50 cents on any 1 dollar bet. 3. If X is Joes average pay off over a year then X = X = . 50 , X = X 104 =1.5497. The CLT says that X is approximately normal. 4. Use the normal approximation so P ( X 1) =pnorm(1,.50,1.5497,lower.tail=F)=0.3734828 Problem 5.60 1. The central limit theorem says that the sample means will be roughly Nor-mal. Note that the distribution of individual scores cannot ahve extreme outliers because all the scores are between 1 and 7. 2. Y has mean 4.8 and standard deviation .2835 X has mean 2.4 and standard deviation .3024 3. Y- X is approximately normal with mean 2.4 and standard deviation .4145 4. P ( Y- X ) = pnorm(1,2.4,.4145,lower.tail=F)=0.9996 2...
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Homework4 - 3) = . 31+ . 40 = . 71 , P ( X 3) =

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