Math 340 Homework Set 8 ___________________________________________________________________________ 1. (p.108, #8) Here we want the P(X=100) where X is binomial with n=600 and p=1/6. The exact value would be found using the formula for binomimal dist which too cumbersome. Thus, we ask for an estimate instead. To find an estimate for the probability of the event (X=100) use a normal dist with mu=np=(600)(1/6) and sigma=sqrt(npq). The area that corresponds to the prob. in question is then between 99.5 and 100.5. Transform these numbers and use the standard normal to find the needed result. ___________________________________________________________________________ 2. (p.108, #11) You should consider 0.3 as the prob. for success in this problem. In order to put the problem in a language that is familiar consider the following: X = the number of hits in n attempts p = P(S) = 0.3 n = 100. Hence, the expected number of hits (successes) in n=100 attempts is mu = (100)(.3) = 30. Note that in this way of setting up the problem X is binomial with n=100 and p=.3. So, now we are asked some questions about X. For example one of the problems can be formulated as: P( the "proportion" of successes in 100 attempts be > 0.310 ) = P( X > 31). Finally, the P(X>31) can be approximated using a normal distribution with mean mu=30 and s.d. sigma = sqrt(100 .3 .7). The mean and the S.D. of X are as follows: mu = n p = (100)(.3) = 30 sigma = \sqrt[n p (1-p)] = \sqrt[(100)(0.3)(0.7)] = 4.58 Now, let Y = N(30, 4.58). Then, we have the following normal approximation to the binomial X = B(100, 0.3): P(X > 31) \approx P(Y > 30.5) = 1 - \Phi(z), where z is the z-score for x = 30.5, and Phi is the cumulative prob dist for the standard normal dist whose values are tabulated in the back of the book. It turns out that z = (30.5 - 30)/4.58 = 0.11 => 1 - \Phi(0.11) = 0.4562 Otherwise parts are worked out likewise. ______________________________________________________________________________ 3. (p.108, #12) We are given that: a. n=10,000 b. X = the number of heads in 10,000 tosses. c. the coin is fair, so p=1/2. d. 2/3 = P(5000-m leq X leq 5000+m), where leq means less than or equal to. To find m note that the following is true: 2/3 = P( 5000-m leq X leq 5000+m ) ~ P( 5000-m leq Y leq 5000+m ) = P( - z leq Z leq z ) where Y is a normal r.v. with the same mean and standard deviation as X, Z is the standard normal distribution, and z is the z-score for 5000+m. Namely, z = (5000+m - mu) / sigma. Hence using the table in the back of the text one can find z and work his/her way backward to find m. ______________________________________________________________________________ 4. (p.108, #13) p = the proportion of the people in a population who intend to vote for a particular candidate. (p is unknown and will remain an unknown throughout) p^ = the proportion of the people in a sample of size n who intend to vote for a particular candidate. (p^ will be known to us as soon as we have interviwed everyone in our sample.) The meaning ofa 95% confidence interval is that: .95 = P( p^ - E < p < p^ + E) = P( - z < Z < z ) Recall that E = (z)(sigma) and sigma = sqrt(npq). In this problem we are told that E = 1 and we are looking for n. If you put all of this information together, you will find that an estimate for n can be obtained inview of the following fact: pq = p(1-p) leq 1/4 for all 0 leq p leq 1. _____________________________________________________________________________ 5. (page 121, #3) In this problem we are given: a. Our experiment consists of rolling 100 dice and counting the number of sixes that occur. b. We perform this experiment n=365 times (once a day for a year). c. Success is the event that the number of occurrences of the number six for one trial (one throw of 100 dice) is 25 or more. d. X = is the number of successes in n=365 trials. e. p = P(S) = 0.022. Hence, X is binomial where success is a rare event and n is large. Therefore, we can approximate the dist of X with that of the corresponding Poisson dist. ____________________________________________________________________________ 6. (p.121, #5) Same as above with n=52 (there are 52 weeks in a year) and p=0.01. X is the number of successes in 52 trials. Thus, P(X=k) \approx e^(-mu) mu^k / k!, where \approx means "is approximately equal to." ____________________________________________________________________________ 7. (p.121, #7) The first thing that we have to note in this problem is the misprint: Change all S's to X's. With that change now we can do the problem. a. m = int(np+p) = int (25 .1 + .1) = int (2.6) = 2, since the int(a) is the greatest integer smaller than or equal to a. b. The exact value of the the prob for m=2 successes = P(X=2) P(X=2) = ("25 choose 2") p^2 q^23 c. For normal approximation. P(X=2) = P(1.5 < Y < 2.5) = P(z_1 < Z < z_2) z_1 = (1.5 - mu) / sigma and z_2 = (2.5 - mu) / sigma where mu and sigma follow the usual formulas for binomial dist. Now use the table in the book to find this prob. mu = np = 2.5 and sigma = \sqrt[np(1-p)] = 1.5. Hence, z_1 = (1.5-2.5)/1.5 = -0.67, and z_2 = (2.5-2.5)/1.5 = 0. Finally, P(X=2) = \Phi(0.67) - \Phi(0) = 0.7486 - 0.5 = 0.2486. The difference with the answer in the back of the book is due to the rounding off of -2/3 by -0.67. d. For Poisson approximation. P(X=2) \approx e^(-mu) mu^2 / 2! e. For a fixed p if n gets larger, a binomial dist can be approximated more accurately by a normal dist. Where as if we let p get smaller as n gets larger, then a Poisson approximation will provide a better result. Consult the histograms on pages 117 and 120.