1 week 3.. 2 3 normal distribution bernoulli trials binomial distribution exponential distribution...
DESCRIPTION
3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTIONTRANSCRIPT
![Page 1: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/1.jpg)
1Week 3.
![Page 2: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/2.jpg)
2Week 3.
![Page 3: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/3.jpg)
3
NORMAL DISTRIBUTIONBERNOULLI TRIALS
BINOMIAL DISTRIBUTIONEXPONENTIAL DISTRIBUTION
UNIFORM DISTRIBUTIONPOISSON DISTRIBUTION
![Page 4: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/4.jpg)
4
note thepoint of inflexion
note thebalance point
![Page 5: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/5.jpg)
5
SD=15
MEAN = 100
point ofinflexion
![Page 6: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/6.jpg)
6
5
50
![Page 7: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/7.jpg)
7
6.3
39.7
![Page 8: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/8.jpg)
8
~68%
Illustrated for theStandard NormalMean=0, SD=1
![Page 9: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/9.jpg)
9
~95%
Illustrated for theStandard normalMean=0, SD=1
![Page 10: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/10.jpg)
10
15
100
~68/2=34%
~95/2=47.5%
130 85
![Page 11: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/11.jpg)
11
15
100
~68/2=34%
~95/2=47.5%
130 85
![Page 12: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/12.jpg)
12
15IQ Z
100
0
1
Standard Normal
![Page 13: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/13.jpg)
13
![Page 14: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/14.jpg)
14
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
![Page 15: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/15.jpg)
15
x p(x)
1 p (1 denotes “success”)0 q (0 denotes “failure”) __ 1
0 < p < 1q = 1 - p
![Page 16: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/16.jpg)
16
P(success) = P(X = 1) = pP(failure) = P(X = 0) = q
e.g. X = “sample voter is Democrat” Population has 48% Dem.
p = 0.48, q = 0.52P(X = 1) = 0.48
![Page 17: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/17.jpg)
17
P(S1 S2 F3 F4 F5 F6 S7) = p3 q4
just write P(SSFFFFS) = p3 q4
“the answer only depends upon howmany of each, not their order.”
e.g. 48% Dem, 5 sampled, with-repl:P(Dem Rep Dem Dem Rep) = 0.483 0.522
![Page 18: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/18.jpg)
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 .482 0.522 ~ 0.374.
There are 6 ways to arrange 2D 2R.
![Page 19: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/19.jpg)
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 .483 0.522 ~ 0.299.
There are 10 ways to arrange 3D 2R.Same as the number of ways to select 3 from 5.
![Page 20: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/20.jpg)
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.
![Page 21: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/21.jpg)
21
Let random variable X denote the number of“S” in n independent Bernoulli p-Trials.By definition, X has a Binomial Distributionand 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.
![Page 22: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/22.jpg)
22
n!/(x! (n-x)!) is the count of how manyarrangments there are of a string of x letters“S” and n-x letters “F.”.
px qn-x is the shared probability of each stringof x letters “S” and n-x letters “F.”(define 0! = 1, p0 = q0 = 1 and the formula goesthrough for every one of x = 0 through n)
is short for the arrangement count =
![Page 23: 1 Week 3.. 2 3 NORMAL DISTRIBUTION BERNOULLI TRIALS BINOMIAL DISTRIBUTION EXPONENTIAL DISTRIBUTION UNIFORM DISTRIBUTION POISSON DISTRIBUTION](https://reader033.vdocuments.net/reader033/viewer/2022051123/5a4d1ade7f8b9ab0599761d0/html5/thumbnails/23.jpg)
23