1. 1 Week 3.

2. 2

3. 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION

4. 4 note the point of inflexion note the balance point

5. 5 SD=15 MEAN = 100 point of inflexion

6. 6 5 50

7. 7 6.3 39.7

8. 8 ~68% Illustrated for the Standard Normal Mean=0, SD=1

9. 9 ~95% Illustrated for the Standard normal Mean=0, SD=1

10. 10 15 100 ~68/2 =34% ~95/2=47.5% 130 85

11. 11 15 100 ~68/2 =34% ~95/2=47.5% 130 85

12. 12 15 IQ Z 100 0 1 Standard Normal

13. 13

14. 14 P(Z > 0) = P(Z < 0 ) = 0.5 P(Z > 2.6 6 ) = 0.5 - P(0 < Z < 2.6 6 ) = 0.5 - 0.4961 = 0.0039 P(Z < 1.9 2 ) = 0.5 + P(0 < Z < 1.9 2 ) = 0.5 + 0.4726 = 0.9726

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

16. 16 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. 17 P(S1 S2 F3 F4 F5 F6 S7) = p 3 q 4 just write P(SSFFFFS) = p 3 q 4 “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) = 0.48 3 0.52 2

18. 18 e.g. P(exactly 2 Dems out of sample of 4) = P(DDRR) + P(DRDR) + P(DDRR) + P(RDDR) + P(RDRD) + P(RRDD) = 6.48 2 0.52 2 ~ 0.374. There are 6 ways to arrange 2D 2R.

19. 19 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) = 10.48 3 0.52 2 ~ 0.299. There are 10 ways to arrange 3D 2R. Same as the number of ways to select 3 from 5.

20. 20 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. 21 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)!) ) p x q n-x e.g. P(44 Dems in sample of 100 voters) = (100!/(44! 56!)) 0.48 44 0.52 100-44 = 0.05812.

22. 22 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.”. p x q n-x is the shared probability of each string of x letters “S” and n-x letters “F.” (define 0! = 1, p 0 = q 0 = 1 and the formula goes through for every one of x = 0 through n) is short for the arrangement count =

23. 23