Raoul LePage Professor STATISTICS AND PROBABILITY stt.msu/~lepage click on STT315_Sp06
23
1 Week 3.
description
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 - PowerPoint PPT Presentation
Transcript of Raoul LePage Professor STATISTICS AND PROBABILITY stt.msu/~lepage click on STT315_Sp06
1Week 3.
2Week 3.
3
NORMAL DISTRIBUTIONBERNOULLI TRIALS
BINOMIAL DISTRIBUTIONEXPONENTIAL DISTRIBUTION
UNIFORM DISTRIBUTIONPOISSON DISTRIBUTION
4
note thepoint of inflexion
note thebalance point
5
SD=15
MEAN = 100
point ofinflexion
6
5
50
7
6.3
39.7
8
~68%
Illustrated for theStandard NormalMean=0, SD=1
9
~95%
Illustrated for theStandard normalMean=0, SD=1
10
15
100
~68/2=34%
~95/2=47.5%
130 85
11
15
100
~68/2=34%
~95/2=47.5%
130 85
12
15
IQ Z
100
0
1
Standard Normal
13
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
15
x p(x)
1 p (1 denotes “success”)0 q (0 denotes “failure”) __ 1
0 < p < 1q = 1 - p
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
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
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.
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.
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
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.
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 =
23
