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

This resource provides solutions for suggested exercises on normal distribution, Bernoulli trials, binomial distribution, exponential distribution, uniform distribution, and Poisson distribution. Explore probability models with diverse applications and understand the mean, standard deviation, and standard scores in these distributions.

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

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

  2. 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.

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

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

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

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

  7. 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

  8. 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%

  9. 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%

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

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

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

  13. STANDARD SCORES CONVERT TO 0 MEAN; SD 1

  14. 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

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

  16. 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

  17. 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

  18. 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.

  19. 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.

  20. 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.

  21. 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.

  22. 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

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