Transcript Chapter 16 Random Variables

Chapter 16
Random Variables
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Random Variable
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Variable that assumes any of several
different values as a result of some random
event. Denoted by X
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Discrete (finite number of outcomes)
Continuous
Probability Model or Probability
Distribution
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Collection of all the possible values and
their corresponding probabilities
Random Variables
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Ex: Insurance Company
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Charges $50 per policy
Death $10000
Disability $5000
Policy holder
outcome
Death
Disability
Neither
Payout
X
10,000
5,000
0
Probability
P(X=x)
1/1000
2/1000
997/1000
Random Variables
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Ex: Insurance Company
Expected Value of a Random Variable (or Mean)
  E( X )   x  P( X  x)
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E(X) =
10,000(1/1,000)+5000(2/1000)+0(997/1000)
Random Variables
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Standard Deviation of the random
variable
  Var( X )  ( x  )  P( X  x)
2
2
Policy holder
outcome
Payout
X
Probability
P(X=x)
Deviation
Death
10,000
1/1000
(10,000-20)
Disability
5,000
2/1000
(5,000-20)
Neither
0
997/1000
(0-20)
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Exercises page 427, just checking 411
More About Means and
Variances
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Shift
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Adding or subtracting a constant from the data
shifts the expected value, but doesn’t change the
variance or standard deviation
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E(X±c) = E(X) ± c
Var(X±c) = Var(X)
Rescaling
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Multiplying each value of a random variable by a
constant multiplies the expected value and the
standard deviation by that constant
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E(aX)= a E(X)
SD(aX) = a SD(X)
Var(aX) = a2 Var(X)
More About Means and
Variances
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Adding or subtracting random variables
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The mean of the sum of two random variables is
the sum of the means
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The mean of the difference of two random
variables is the difference of the means
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E(X+Y) = E(X) + E(Y)
E(X-Y) = E(X) - E(Y)
If the random variables are independent, the
variance of their sum or difference is always the
sum of the variances
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Var(X±Y) = Var(X) +Var(Y)
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Just checking 418
Continuous Random Variables
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We don’t have discrete outcomes, the
random variable can take on “any” value.
Example
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Packaging Stereos
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Packing
Boxing
Normal
Normal
E(P)=9
E(B)=6
SD(P)=1.5
SD(B)=1
What is the probability that packing two consecutive
systems takes over 20 minutes?
What percentage of the stereo systems take longer to
pack than to box?
Chapter 17
Probability Models
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Bernoulli Trials
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Only two possible outcomes
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The probability of a success denoted “p” is the
same on every trial.
The trials are independent
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Success
Failure
If this assumption is violated, it is still ok to proceed as
long as the sample is smaller than 10% of the
population.
Ex: Coin toss, rolling a die ?
Geometric probability model
for Bernoulli trials: Geom (p)
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p = probability of success
q =1-p probability of failure
X = number of trials until the first success
occurs
P(X=x)=qx-1p
Expected Value

Standard deviation
1
p

q
p2
The Binomial Model
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Ex:
p=1/6
q=5/6
Probability of 2 successes in 5 trials?
(5/6)3(1/6)2
What about the other orders?
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The number of different orders in which we can have
k successes in n trials is written nCk and pronounced
“n choose k” (The C actually stands for
combinations)
n!
n Ck 
k!(n  k )!
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Where n!=1 x 2 x 3 x … x (n-1) x n
Binomial probability model for
Bernoulli trials: Binom (n,p)
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n = number of trials
p = probability of success
q = 1 – p probability of failure
K = number of successes in n trials
P(K=k)= nCk pkqn-k
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where
n Ck 
n!
k!(n  k )!
  np
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Mean
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Standard deviation
  npq
Exercises
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Step-by-step page 435 and 438