Raoul LePage

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
SD15
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 Mean0, SD1
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 Mean0, SD1
95
10
IQ DISTRIBUTION NORMAL, MEAN 100 STANDARD
DEVIATION 15
15
68/2 34
95/247.5
130
85
100
11
IQ DISTRIBUTION NORMAL, MEAN 100 STANDARD
DEVIATION 15
15
68/2 34
95/247.5
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 gt 0) P(Z lt 0 ) 0.5 P(Z gt 2.66) 0.5 -
P(0 lt Z lt 2.66) 0.5 -
0.4961 0.0039 P(Z lt 1.92) 0.5 P(0 lt Z lt
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 lt p lt 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
(11 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
23