Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
I ADDED THE WORK IN RED, ASK ME IF you have doubts A GOOD BONUS AS THE LAST TIME IS WELCOME 1. In a recent year, the mean number of strokes per hole for a famous golfer was approximately 3.8 (a). Find the variance and standard deviation using the fact that variance of a Poisson distribution is ơ²=µ If x has a poisson distribution with mean = u, Variance (x) = u (Mean = variance) And always standad deviation = sd = V(x) (for every distribution) Answer: Variance V(x) = u = 3.8, sd = √V(x) = √3.8 = 1.949 (b) How likely is this golfer to play an 18-hole round and have more than 72 strokes. - The variance is: 3.8 - The standard deviation is: 1.949 - Find the mean number of strokes for 18 holes if the mean number of strokes per hole is 3.8 If you have a poisson distribution X (X counts the ocurrences in a unit of time) with mean = u per unit of time then If Y is a random variable that counts the ocurrences in n units of times then Y has a Poisson distribution with mean = nu Mean number of strokes for 18 holes = nu = 18(3.8) = 68.4 - The probability of more than 72 strokes in an 18-hole round is: Let X = strokes for 18 holes X has a poisson distribution with mean 68.4 P(X = x ) = (e-68.4 68.4x)/x! P(X>72) = 1- P(X≤72) = 1-[P(X=0)+P(X=1)+.....+P(X=71)+P(X=72)] = Since you have to add 73 terms (from 0 to 72) you need a software for comuting this In excel you will find the poisson distribution and you can compute P(X≤x) easily Syntax : Poisson (x, mean,cumulative) Cumulative can be TRUE or FALSE You have to use TRUE if you want to compute P(X≤x) You have to use FALSE if you want to compute P(X = x) In this case we want to find P(X>72) = 1- P(X≤72) To compute P(X≤72) since x=72, mean = 68.4 put POISSON ( 72, 68.4,TRUE) = 0.695 Then P(X>72) = 1- P(X≤72) = 1-0.695 = 0.305 Answer: 0.305 2. 33% of adults say cashews are their favorite kind of nut. You randomly select 12 adults and ask each to name his or her favorite nut. Find the probability that the numbers who say cashews are their favorite nut is Let X = number of people that prefer cashews X has a binomial distribution with n=12 and p = 0.33 (a) exactly three, P(X=3) = (12C 3)(0.33)^3(0.67)^9 = 0.215 (rounded to 3 dp) Answer: 0.215 (b) at least four, and P(X≥4) = 1-P(X≤3) To compute P(X≤3) you can use excel too See this page: http://homepages.wmich.edu/~bwagner/StatReview/Binomial/binomia%20in%20excel.htm Then: P(X≤3) = BINOMDIST(3;12;0.33;TRUE) = 0.403 (rounded to 3 dp) P(X≥4) = 1-P(X≤3) = 1-0.597 = 0.403 Answer: 0.597 (c) at most two. P(X≤2)= 0.188 (see part b) Answer: 0.188 3. a. construct a binomial distribution n= 6 p= 0.36 b. Find the probability of each value of x to construct the binomial distibution You must use: P(X = x ) = (nCx)p^x(1-p)^(n-x) (or use Excel, see exercise 2) x P(x) (By hand ) EXCEL 0 (6C0)(0.36)^0(0.64)^6 = 0.069 BINOMDIST(0;6;0.36;FALSE) 1 (6C1)(0.36)^1(0.64)^5 = 0.232 BINOMDIST(1;6;0.36;FALSE) 2 (6C2)(0.36)^2(0.64)^4 = 0.326 BINOMDIST(2;6;0.36;FALSE) 3 (6C3)(0.36)^3(0.64)^3 = 0.245 BINOMDIST(3;6;0.36;FALSE) 4 (6C4)(0.36)^4(0.64)^2 = 0.103 BINOMDIST(4;6;0.36;FALSE) 5 (6C1)(0.36)^5(0.64)^1 = 0.023 BINOMDIST(5;6;0.36;FALSE) 6 (6C0)(0.36)^6(0.64)^0 = 0.002 BINOMDIST(6;6;0.36;FALSE) (Do not round until the final answer Then round to th nearest thousandth as needed) c) Find the probability that exactly two people will name “talking on cell phones.” P(2)= 0.326 (Just see the table above for x = 2) (Round to the nearest thousandth as needed) 4. Assume the geometric distribution applies. Use the given probability of success p to find the indicated probability. P(X =x) = (1-p)^(x-1) p Find P(4) when p=0.79 P(4) = 0.21^3(0.79) = 0.007 Answer: 0.007 5. Given that x has a Poisson distribution with µ=4, what is the probability that x=7 P(x=7) = (e-447)/7! = 0.0595 Answer: 0.0595 6. Answer: B) Binomial 7. (a) Find the probability that the first warped glass item is the 10th item produced. Using the formula in exercise 4, p=1/500, x=10, 1-p= 499/500 P(10) = (499/500)^9 (1/500) = 0.00196 Answer: 0.00196 (b) Find the probability that the first warped glass item is the first, second, or third item produced. Here we have to compute P(X≤3) = P(X=1)+P(X=2)+P(X=3) (there is no excel function) P(X=1) = 1/500 P(X=2) = (499/500)(1/500) P(X=3)= (499/500)^2(1/500) P= 0.006 (Round to the nearest thousandth as needed) c) Find the probability that none of the first 10 glass items produced are defective. P= (1-1/500)^10 = 0.980 P= 0.980 Round to the nearest thousandth as needed 8. Answer: B (Binomial)