Raoul LePage Professor STATISTICS AND PROBABILITY stt.msu/~lepage click on STT315_Sp06

Raoul LePage Professor STATISTICS AND PROBABILITY www.stt.msu.edu/~lepage click on STT315_Sp06 Week 3.

suggested exercises solutions given in text 3-33, 3-41, 3-42 (except b, c, h, m, n), 3-43, 3-49, 3-57 (except c, d), 3-59, 3-61, 3-63, 3-65. textbook exercises are not comprehensive this chapter Week 3.

NORMAL DISTRIBUTIONBERNOULLI TRIALSBINOMIAL DISTRIBUTIONEXPONENTIAL DISTRIBUTIONUNIFORM DISTRIBUTIONPOISSON DISTRIBUTION PROBABILITY MODELS HAVING BROAD APPLICATION

NORMAL DISTRIBUTION: WHERE ARE THE MEAN AND STANDARD DEVIATION IN THIS PICTURE? note the point of inflexion note the balance point

IQ DISTRIBUTION: ~NORMAL, MEAN 100 STANDARD DEVIATION 15 point of inflexion SD=15 MEAN = 100

DISTRIBUTION OF THE NUMBER OF HEADS IN 100 COIN TOSSES: APPROXIMATELY NORMAL, MEAN 50, STD DEVIATION 5 5 50

DISTRIBUTION OF THE NUMBER OF ACCIDENTS IN ONE MONTH IF WE AVERAGE 39.7 PER MONTH: APPROXIMATELY NORMAL, MEAN 39.7, STD DEVIATION 6.3 6.3 39.7

NORMAL DISTRIBUTIONS ARE ALIKE IN SD UNITS FROM THE MEAN ~ 68% WITHIN 1 SD OF MEAN ~ 95% WITHIN 2 SD OF MEAN Illustrated for the Standard Normal Mean=0, SD=1 ~68%

NORMAL DISTRIBUTIONS ARE ALIKE IN SD UNITS FROM THE MEAN ~ 68% WITHIN 1 SD OF MEAN ~ 95% WITHIN 2 SD OF MEAN Illustrated for the Standard normal Mean=0, SD=1 ~95%

IQ DISTRIBUTION: ~NORMAL, MEAN 100 STANDARD DEVIATION 15 15 ~68/2 =34% ~95/2=47.5% 130 85 100

STANDARD SCORES CONVERT TO 0 MEAN; SD 1 IQ Z 1 15 0 Standard Normal 100

STANDARD SCORES CONVERT TO 0 MEAN; SD 1

Z - TABLE CUT AND PASTE P(Z > 0) = P(Z < 0 ) = 0.5 P(Z > 2.66) = 0.5 - P(0 < Z < 2.66) = 0.5 - 0.4961 = 0.0039 P(Z < 1.92) = 0.5 + P(0 < Z < 1.92) = 0.5 + 0.4726 = 0.9726

BERNOULLI DISTRIBUTION x p(x) p (1 denotes “success”) 0 q (0 denotes “failure”) __ 1 0 < p < 1 q = 1 - p

Notation: BERNOULLI RANDOM VARIABLE X P(success) = P(X = 1) = p P(failure) = P(X = 0) = q e.g. X = “sample voter is Democrat” Population has 48% Dem. p = 0.48, q = 0.52 P(X = 1) = 0.48

INDEPENDENT BERNOULLI-p "S" denotes success "F" denotes failure P(S1 S2 F3 F4 F5 F6 S7) = p3 q4 just write P(SSFFFFS) = p3 q4 “the answer only depends upon how many of each, not their order.” e.g. 48% Dem, 5 sampled, with-repl: P(Dem Rep Dem Dem Rep) = 0.483 0.522

BINOMIAL DISTRIBUTION FOR THE TOTAL NUMBER OF SUCCESSES IN INDEPENDENT p-BERNOULLI TRIALS. e.g. P(exactly 2 Dems out of sample of 4) = P(DDRR) + P(DRDR) + P(DDRR) + P(RDDR) + P(RDRD) + P(RRDD) = 6 .482 0.522 ~ 0.374. There are 6 ways to arrange 2D 2R.

BINOMIAL DISTRIBUTION FOR THE TOTAL NUMBER OF SUCCESSES IN INDEPENDENT p-BERNOULLI TRIALS. e.g. P(exactly 3 Dems out of sample of 5) = P(DDDRR) + P(DDRDR) + P(DDRRD) + P(DRDDR) + P(DRDRD) + P(DRRDD) + P(RDDDR) +P(RDDRD) + P(RDRDD) + P(RRDDD) = 10 .483 0.522 ~ 0.299. There are 10 ways to arrange 3D 2R. Same as the number of ways to select 3 from 5.

COUNTING ARRANGEMENTS 5! ways to arrange 5 things in a line Do it thus (1:1 with arrangements): select 3 of the 5 to go first in line, arrange those 3 at the head of line then arrange the remaining 2 after. 5! = (ways to select 3 from 5) 3! 2! So num ways must be 5! /( 3! 2!) = 10.

BINOMIAL FORMULA Let random variable X denote the number of “S” in n independent Bernoulli p-Trials. By definition, X has a Binomial Distribution and for each of x = 0, 1, 2, …, n P(X = x) = (n!/(x! (n-x)!) ) px qn-x e.g. P(44 Dems in sample of 100 voters) = (100!/(44! 56!)) 0.4844 0.52100-44 = 0.05812.

Caveats: Binomial n!/(x! (n-x)!) is the count of how many arrangments there are of a string of x letters “S” and n-x letters “F.” . px qn-x is the shared probability of each string of x letters “S” and n-x letters “F.” (define 0! = 1, p0 = q0 = 1 and the formula goes through for every one of x = 0 through n) is short for the arrangement count = Binomial Coefficient

Raoul LePage Professor STATISTICS AND PROBABILITY stt.msu/~lepage click on STT315_Sp06