Week 3.
1
Week 3.
2
NORMAL DISTRIBUTION
BERNOULLI TRIALS
BINOMIAL DISTRIBUTION
EXPONENTIAL DISTRIBUTION
UNIFORM DISTRIBUTION
POISSON DISTRIBUTION
3
note the
point of
inflexion
note the
balance point
4
SD=15
MEAN = 100
point of
inflexion
5
5
50
6
6.3
39.7
7
Illustrated for the
Standard Normal
Mean=0, SD=1
~68%
8
Illustrated for the
Standard normal
Mean=0, SD=1
~95%
9
15
~68/2
=34%
~95/2=47.5%
85
100
130
10
15
~68/2
=34%
~95/2=47.5%
85
100
130
11
IQ
15
100
1
Z
0
Standard Normal
12
13
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
14
x
p(x)
1
0
p
q
__
1
(1 denotes “success”)
(0 denotes “failure”)
0<p<1
q=1-p
15
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
16
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 17
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.
18
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. 19
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.
20
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.
21
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
=
22
23