Question

2. Model : We will replace our data with a probability mass function. In other words assume we know that of all the world's beers 70% are low alcohol, 20% are medium alcohol and 10% are high alcohol. We also assume all low alcohol beers are 5.13% ABV, medium are 6.85% ABV and high is 10.20% ABV. Let X be a random variable representing the ABV of a randomly selected beer. a. (4pts) Draw a cumulative distribution function for X. Doesn't need to be perfect, just do the best you can with the information listed above. b. (3pts) What is the expected value of X? You can label itu. C. (4pts) What is the variance of X? d. (4pts) Assume each can of beer has 12ozs. What is the expected value and variance of the amount of alcohol in ounces per can e. (4pts) Assuming your answer to (d) is correct what is the expected value and variance of the amount of alcohol in 6 randomly selected beers. f. (4pts) Assuming your answer to (d) is correct what is the expected value and variance of the amount of alcohol in a randomly selected 6 pack', i.e. each beer of the 6 beers is the same type g. (4pts) Assuming your answer to (d) is correct if you pour 1 randomly selected beer in cup A and another two randomly selected beer in cup B what is the

can you give the Answers to E,F,G,H?

Answer 1

Solution & (since there answer first H). are more than # part i will dist bunīlion plans cornmunication dist bundlion 5:13% 0.40 10. 20 0.90 6:85% 10. 204. 1.00 €(a)= 3x p(a) ECO) = 5.13%. + 0.70 +6.58% * 0.20 +10. 20%, *.10 ECH) = U-5.927% war (2) > € (-) / (N) 0.5 ELN (5.13-5,727) + (6.58-5.927) + (10.20-5.927) (137. = 23838

- to get amount of mean and variance of al chord Content multiply volume by mean and variance ab? percentages. mean alchohool Content per Can - 1275.9244 ounces = 0.71124 onces 3838 variance of alchohol content per can Ounces - 0.286056 ounces ps ( please up vote if you find the answer Satisfactory).