Class Notes (838,799)
Canada (511,099)
Statistics (297)
STAB22H3 (239)
Ken Butler (34)
Lecture

stab22 lecture notes (forgot which one)

22 Pages
130 Views
Unlock Document

Department
Statistics
Course
STAB22H3
Professor
Ken Butler
Semester
Fall

Description
STAB22, LEC#17 CHAPTER 17 (COVERS 17,18) Reminder: BT [211] (ex) of BT (Bernoulli Trials) using "Tossing a coin" example 1. two possible outcomes: "success" (p) or "failure" (q) - let success be event we get heads - let failure be event we get tails 2. p doesn't change - chance of getting success (ex. subsequent head) doesn't change from one trial to the next 3. trials are indep. - if you get 6 heads, that doesn't chance chance of getting tails We use BT in one of two ways, dep. on what is of interest: 1. want to know #trials until first success - aka: how long do we have to keep going until first success? => in this case, use Geometric model 2. want to know #successes in fixed #trials - we have fixed # of attempts, so want to know how many times we succeed in these attempts => in this case, use Binomial model [212] Binomial model - how many attempts will be made (n), how likely are you to succesed on each one (p), and how many successes u like to see (k) BINOMIAL TABLE (THIS WILL BE PROVIDED TO US ON EXAM) n k p 0.15 0.20 8 0 0.2725 0.1678 8 1 0.3847 0.3355 .... ^- above is layout of table, along with two rows of data. For all questions on this slide, we are using: n = 10, p = 0.15 (ex) coffee cups - buy 10 cups (n = 10), and 15% chance of being winner on any cup of cofee 1. What is P of 0 prizes? (k=0) - that is telling us p = 0.15, n =10, k = 0, so our p of interest = 0.1969 - p =0.15, n = 10, x = 0 successes - this is chance of getting exactly no successes at all up to 10, if chance of succeeding is 0.15 - this is chances of getting no successes at all => (these q's simply req. looking up on table) 2. What is P of exactly 1 prize? (k=1) 0.3474 3. What is P of exactly 2 prizes? (k=2) 0.1937 4. What is P of getting 2 prizes or less? - this is asking (P of no prizes) + P(exactly 1) + P(exactly 2) - 0.1969 + 0.3474 + 0.1937 = 0.7380 => about 74% chance of winning 2 or less prizes (we are literally adding up the answers we got from Q's 1-3) - its complement, "NOT 2 prizes or less", is 5. What is P of getting 2 prizes or more? - we can do: P(2) + P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) + P(10) = OR - we can get the complement: P(1 prize or less) and then subtract this from 1 to get P(2 prizes or more) - the P of getting 2 prizes or more means that you are NOT getting 1 prize or less (ie. less than 2 prizes) => 1 - (0.1969 + 0.3474) = 0.4557 - what we did here was Notice: the same thing as saying "3 prizes or more". - the bottom - n=10, k=10, => in the context of our problem, this means that it is v.unlikely (~0%) to get all 10 prizes, out of 10 trials, given that chance of success is 0.15. [213] WHAT IF P> 0.5 IN BINOMIAL TABLE? - Binomial table does not go p > 0.5 - Suppose n=8 and p=0.7, What is the probability of: 1- exactly 7 successes? 2- 7 or more successes? - Idea: count failures instead of successes ------- 1. P(success) = 0.7=> this means: P(fail) = 0.3 If there are 8 trials, and we have 7 successes, then => this means 1 failure. (we converted our success info into failure info) - so now, we use table with: - p = 0.3 - k = 1 - n = 8 - b/c we equivalently expressed our "success" info in terms of "failure", and now we have a p we can work with that is p < 0.5, so we can use binomial table. - chances of getting exactly 7 successes => this is answer we want; P = 0.1977 We do not have to take this number away from 1 - we aren't taking away from 1, b/c we already turned eth around: #successes -> #failures, and P success -> P failure 2. - 7 or more successes = 7 or 8 successes - P(success) = 0.7 => P(failure) = 0.3 - 7,8 successes => 1,0 failures - if we had either 7 or 8 succeses, then this is equivalent to saying that we had 1 or no failures at all (one failure or less). - so we have to get: a) p = 0.3, k = 1, n = 10 P(exactly 1 failure) = 0.1977 b) p = 0.3, k = 0, n = 10 P(no failure) = 0.0576 , and then, add these two up. => P(7 or more successes) = 0.1977 + 0.0576 = 0.2553. Implication -successes and failures are kind of symmetric in Binom distrib. [214] MEAN AND SD OF BINOMIAL DISTRIBUTION - Mean of number of successes in binomial distribution (n)(p) - SD of number of successes = √ aka √ For derivation of SD formula, see MATH BOX, p451 (did calc. for prev ex.; n = 10, p =0.15) ---- EXAMPLE (using n = 10, p =0.15) √ √ - 10 successes in possible, but v.unlikely - so distribution likely to be skewed to right USE STATCRUNCH TO GET PROBABILITY HISTOGRAMS OF BINOMIAL DISTRIBUTIONS (STA, CALCULATORS, BINOMIAL) Figure 1 - n=10, p = 0.15 - R-skewed distribution Figure 2 - n = 10, p = 0.75 - L-skewed distribution [216] Figure 3 - n = 100, p = 0.15 - looks symmetric, but tiny bit skewed to right Figure 4 - n = 10, p = 0.50 - shape of distribution is symmetric - works even for small number of trials (ie. when n is small) HOW DOES THE SHAPE DEPEND ON P? - if p < 0.5, R-skewed - if p > 0.5, L-skewed - if p = 0.5, symmetric WHAT HAPPENS TO THE SHAPE AS n INCREASES? - shape becomes more normal - - if #trials is large, even if p is far away from 0.5 (ex. p =0.15), then shape becomes normal, as long as #trials is big enough WHAT DOES THIS SUGGEST TO DO IF n IS TOO LARGE FOR THE TABLES? - Binomial table will probably go up to n = 20 - if n > 20 and we do not have that "n" given on the table, then cannot calculate exactly - but we know that once it becomes as big as 100, even for v.small p, shape becomes normal-like [217] Given that n is too large (n is too big for table), then try NORMAL APPROXIMATION TO BINOMIAL - we get mean and SD of binomial using Binomial model formula, but then we assume that it is normal - having done so, we can use z-score formula (Example) - P(10 or fewer prizes in 100 coffee cups) = ? - n = 100, recall: p = 0.15 (from prev slides) - Using this, we can calculate mean and SD: - - √ - then, for 10 prices (y), - we use mean be mu = 15, SD be sigma = 3.57, and x be 10 - then, - look up on Table Z (in textbook) for proportion less, which is 0.0808. - remember: this is an APPROXIMATION. => this is saying that the probability that we win 10 or fewer prizes after buying 100 coffee cups is about less than 9% - however, Professor Butler did it with MyStatCrunch and found exact answer to be 0.0994 WORKS IF a) n LARGE AND b) p NOT FAR FROM 0.5 => req. that #trials be big, and p not be far from 0.5, and when we multiply n and p together, we get a product that is larger than 10. - then, approximation would be valid. - the bigger n gets, the more normal it gets - the closer p is to 0.5, the more normal it will be - if p is long way away from 0.5, then need bigger
More Less

Related notes for STAB22H3

Log In


OR

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


OR

By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.


Submit