# STAT 202 Lecture Notes - Lecture 1: Random Variable, Square Root, Probability Distribution

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Published on 26 Sep 2016

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Stat 202-

September 19, 2016

Square root= [

Probability distribution- chp 6&7

A random variable x can take values 2

If x is discrete, then there are accountable number of values x that can be taken if x is

continuous, then there are an uncountable number of values that can be taken

-for any two values x1 and x2 there is always another value x3 in between

Let x be a discrete random variable (RV) with n possible values

Then the expected value (mean) of x is E(x)=U=n, i=I,Xi,p(xi)

The variance of x is sigma^2 = sum,n,i=I(xi-u)^2p(xi)

The standard deviation = sigma = square root sigma^2

Ex) Flip 3 coins {HHH, HTT, THT, TTH, HHT, HTH, THH, TTT} each has a probability of 1/8, let x be

the number of heads

p(x=0) = 1/8

p(x=1) = 3/8

p(x=2) = 3/8

p(x=3) = 1/8

E(x)=u=0*1/8+1*3/8+2*3/8+3*1/8=1.5

Sigma^2=(0-1.5)^2*1/8+(1-15)^2*3/8+(2-1.5)^2*3/8+(3-1.5)^2*1.8=3/4

Sigma=[3/4 = ([3)/2 ~0.866

Let x be a discrete RV

The PDF (probability distribution function) gives the probability that x takes on a particular

value = p(X=x)

The CDF gives cumulative probabilities, p(X<_x)

Let x be a cont. RV

The PDF(f(x)) can be used (with some calculations to find the prob of being in some interval

The CF of x gives cumulative probabilities, p(X<_x)

Uniform Distribution-

Can be discrete or continuous

The probability of “events” are all the same

Discrete Unit RV X-

X takes on integer values over [a,b]

The probability of each of these values is the same

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PDF: p(X=x)=1/b-a+1

CDF: p(X<_x)=x-a+1/b-a+1

Mean = a+b/2

Variance = ((b-a+1)^2 – 1)/12

Ex) Roll a single di : [a,b]=[1,6]

p(X=x) = 1/6-1+1 = 1/6

p(x<_4) = 4-4+1/6-1+1 = 4/6

Mean = 1+6/2 = 3.5

Std.dev= [((6-1+1)^2 – 1)/12 = [35/12

Let X be a continuous uniform RV

- X can take on any value in the interval [a,b]

- The prob. that X lies in the interval [y,z1] is equal to the prob. that X lies in another

interval [y2,z2] if z1-y1 = z2-y2

Mean = a+b/2

Std.dev= [(b-a)^2/12

CDF= p(X<_x) = x-a/b-a

Ex) Let X be a continuous and uniformly dist. over [0,4]

p(0<_x<_1) = p(2<_x<_3) = p(1.5<_x<_2.5) = ¼

Poisson Distribution-

Models the number of events that occur per time unit

To find probabilities, need one parameter, lamda, which is the average number of events per

time unit

P(X=x) = lamda^x * e^-lamda/x!

X! is “x-factorial”

X! = X(X-1)(X-2)…2-1

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