# 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
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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