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by: Aria Sivick

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# MAT 121 WEEK 9 NOTES MAT 121

Aria Sivick
Syracuse
GPA 3.78

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MAT 121 Week 9 Notes! Enjoy and have a good rest of the weekend!
COURSE
Probability and Statistics for the Liberal Arts I
PROF.
TYPE
Class Notes
PAGES
10
WORDS
KARMA
25 ?

1 review
Nicole

## Popular in Mathematics (M)

This 10 page Class Notes was uploaded by Aria Sivick on Sunday November 1, 2015. The Class Notes belongs to MAT 121 at Syracuse University taught by in Fall 2015. Since its upload, it has received 40 views. For similar materials see Probability and Statistics for the Liberal Arts I in Mathematics (M) at Syracuse University.

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## Reviews for MAT 121 WEEK 9 NOTES

-Nicole

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Date Created: 11/01/15
10/26 The mean (expected value) of the probability distribution of a random variable x is  The variance of the probability distribution of a random variable x is And the standard deviation is Player bets \$1 on red in roulette. Let x = his net winnings. The Binomial Distributions An experiment has two outcomes, SUCCESS and FAILURE (we will count the SUCCESSES), Each time this experiment is performed, P(SUCCESS) = P, P(FAILURE) = 1- P = q.  Do this n times and let x = # of SUCCESSES  Arandom variable x is deﬁned ion this way is said to have a binomial distribution  with parameters n and P. The possible values of x are 0,1,2,…..,n Notation: Whenever there are p and q m a stats problem, p+q = 1 q = 1 - P p + q = 1 p1 + q = 1 10/28   A2 outcome experiment with P(success) = p and P(Failure) = q = 1 - p) is performed n times and x is the number of timed success happens. Such an x has a binary probability distribution with parameters n and p. The possible values of x are 0,1,….,n. Binomial n = 5 P(success) = P What’s P(2)? P(SSFFF) = p^3, q^2 P(FFFSS) = p^3, q^2 1.)Any way of getting 2 successes has probability p^3, q^2 = p^3, q^2. 2.)  There are5C2 = 10 ways to get 2 successes. 1.) SSFFF 2.) SFSFF 3.) SFFSF 4.) SFFFS 5.) FSSFF 6.) FSFSF 7.) FSFFS 8.) FFSSF 9.) FFSFS 10.) FFFSS ____  ____  ____  ____  ____ 1st     2nd    3rd   4th    5th 5C2 = 10 3.) What’s P(4)? Any way of getting 4 successes has probability p^4, q^1 And there are  5C 4= 5 ways Prob: p.p.p.p.q 1.) 2.) 3.) 4.)  If x is a binomial random variable with n trials and p probability of success, then for x = 0,1,2,….,n. P(x) = nC x   p^x,q^n-x Ex.) G.E. 60 - watt bubbles have a probability of failing within one year of .2   . If you have 10 such lightbulbs in your house what are the probabilities that in the next year,  1.) 4 will fail? =Ans is .88 (tableA-1) 2.) 3 will fail? =Ans is .201  3.) at least 3 will fail? P(at least 3 fail) = 1 - P(1) + P(2)                                                             = 1 - (107 + .268 + .302) = 1 - .677 = .323  Its binomial with n = 10 and p = .2  What if it was 17 light bulbs? What’s the probability n = 17, p = .2, x = 4 P(4) = 17C 4(.2)(.8) Calculator = binompdf(17, .2, 4) = .209 10/30

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