1 / 8

MSV 5: DRVs from a Bag

www.making-statistics-vital.co.uk. MSV 5: DRVs from a Bag. Place all the numbers in the bag into the grid below in some order (no repeats!) so that you make a viable discrete probability distribution. How many possible distributions can you make?. Can you find a pair of these

binta
Télécharger la présentation

MSV 5: DRVs from a Bag

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. www.making-statistics-vital.co.uk MSV 5: DRVs from a Bag

  2. Place all the numbers in the bag into the grid below in some order (no repeats!) so that you make a viable discrete probability distribution.

  3. How many possible distributions can you make? Can you find a pair of these that have the same mean and the same variance? The spreadsheet below might be helpful... DRVs from a Bag Spreadsheet Hyperlink http://www.s253053503.websitehome.co.uk/ msv/msv-5/msv-5.xlsx Hint: if you need it! The shared mean is 209/375.

  4. MSV 5: Answers An exercise designed to ensure that students know how to find E(X) and Var(X) for a discrete random variable, and that the probabilities in a probability distribution will always add to one. It is a help in comparing the fractions to put them over a common denominator. So the fractions (1/25, 2/15, 1/5, 3/5, 2/3, 19/25) become (3/75, 10/75, 15/75, 45/75, 50/75, 57/75).

  5. We need triples from the bag that will add to 1. The only possibilities are (1/25, 1/5, 19/25), and (2/15, 1/5, 2/3). These generate six distributions each, giving twelve possibilities in all. To find the distributions with the same mean and variance, we have

  6. Suppose we use the hint. Knowing the mean is 209/375 in both cases gives 285a + 75b + 15c = 209, and 250d + 75e + 50f = 209. Now we can simply check the possibilities. For the right-hand equation, 250d is 190 or 150 or 10, 75e is 3 or 45 or 57, and 50f is 2 or 30 or 38. The only combination that will add to 209 is 150 + 57 + 2. We can use similar logic for the left-hand equation.

  7. So the answers are (with shared variance 6344/140625):

  8. www.making-statistics-vital.co.uk is written by Jonny Griffiths hello@jonny-griffiths.net

More Related