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2Week 3.

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NORMAL DISTRIBUTIONBERNOULLI TRIALS

BINOMIAL DISTRIBUTIONEXPONENTIAL DISTRIBUTION

UNIFORM DISTRIBUTIONPOISSON DISTRIBUTION

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note thepoint of inflexion

note thebalance point

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SD=15

MEAN = 100

point ofinflexion

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5

50

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6.3

39.7

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~68%

Illustrated for theStandard NormalMean=0, SD=1

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~95%

Illustrated for theStandard normalMean=0, SD=1

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15

100

~68/2=34%

~95/2=47.5%

130 85

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15

100

~68/2=34%

~95/2=47.5%

130 85

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15IQ Z

100

0

1

Standard Normal

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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

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x p(x)

1 p (1 denotes “success”)0 q (0 denotes “failure”) __ 1

0 < p < 1q = 1 - p

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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

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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

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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.

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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.

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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.

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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.

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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 =

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