Random Variables

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Random Variables
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Random Variables

• What youll learn
• What is a Random Variable?
• Discrete vs Continuous
• How to construct a valid probability distribution
• Using the information in a probability
  distribution to answer probability questions

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Random Variable

• A Random Variable is a variable whose value is
  numerical outcome of a random phenomenon.
• A Random phenomenon has outcomes that we cannot
  predict, but in the long run has a regular
  distribution.
• Random Variables are either classified as
  Discrete or Continuous

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Discrete vs Continuous

• Discrete RV a discrete rv is a variable whose
  outcomes have a countable number of outcomes.
• In other words, the outcomes can only take on a
  finite number of possibilities
• Examples
• Dice If you roll a six-sided die, the
  possibilities are 1,2,3,4,5, or 6.
• Number of As, Bs, Cs, Ds and Fs in a
  classroom

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Continuous Random Variable

• Continuous
• A continuous variable is a variable that takes on
  all values within a specified interval
• For Example
• Height only the precision of the measurement
  instrument dictates the number of decimal places
• Weight again instrument precision
• Time another continuous random variable

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Discrete Probability Distributions

• A discrete random variable has a countable number
  of outcomes.
• The probability distribution for a discrete rv
  lists each of these outcomes and the probability
  for each outcome

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Discrete RV Probability Distribution

• Requirements for a VALID probability distribution
• Each individual probability must be a number
  between 0 and 1
• The sum of all the probabilities must equal 1

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Discrete Probability Distribution Example

• Consider for a moment a large statistics class.
• Over the years we know that the distribution of
  grades has been 15 each of As and Ds, 30 each
  of Bs and Cs and 10 Fs.
• Consider choosing a student at random from this
  class.

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Creating a Probability Distn

• We can find the probability of the chosen
  students grade by creating the distribution.
• Consider the students grade on a 4-point scale.
  A4, B3, ect.

Grade 0 1 2 3 4
Probability 0.10 0.15 0.30 0.30 .015
10
Grade 0 1 2 3 4
Probability 0.10 0.15 0.30 0.30 0.15

• Verify the distribution is valid
• All probabilities are between 0 and 1
• Sum of the probabilities 1
• .10 .15 .30 .30 .15 1
• Since both conditions have been met, this is a
  valid probability distribution

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Grade 0 1 2 3 4
Probability 0.10 0.15 0.30 0.30 0.15

• Now lets use the distribution to find some
  probabilities
• Find the probability that the student got a B in
  the class
• P(X3) 0.30 30
• We can read this directly from the table

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Grade 0 1 2 3 4
Probability 0.10 0.15 0.30 0.30 0.15

• Find the probability that the student got a C or
  better.
• P(x 2) P(x2) P(x3) P(x4)
• .30 .30 .15 .75 75
• To find this probability we find each individual
  probability and find the sum.

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Grade 0 1 2 3 4
Probability 0.10 0.15 0.30 0.30 0.15

• Find the probability that the student got less
  than a B
• P(xlt 3) P(x2) P(x1) P(x0)
• .30 .15 .10 .55 55
• Notice that since we want strictly less than 3,
  we only find the sum from 2 down.

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Inequality Signs

• Lets recall what signs to use when we are
  looking for inequalities
• Less than lt
• At Most
• Greater than gt
• At least
• Remember if the equal sign is included, we
  include the probability at the endpoint.
• If a strict inequality, we do not include the
  probability at the end point.

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Finding the Mean and Standard Deviation

• Consider
• As the head of inventory for Knowway computer
  company, you were thrilled that you had managed
  to ship 2 computers to your biggest client the
  day the order arrived. You are horrified,
  though, to find out that someone restocked
  refurbished computers in with the new computers
  in your storeroom. The shipped computers were
  selected randomly from the 15 computers in stock,
  but 4 of those were actually refurbished.
• If your client gets 2 new computers, things are
  fine. If the client gets a refurbished computer,
  it will be sent back at your expense--100and
  you can replace it. However, if both computers
  are refurbished, the client will cancel the order
  this month and youll lose 1000. Whats the
  expected value and the standard deviation of your
  loss?

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Mean and Standard Deviation

• The first thing we need to do is find the
  probability distribution for the situation.
• Lets define our random variable as the amount of
  loss that occurs for the company.
• The values that X can take are
• 0 100 1000
• (2N) (1N/1R ) (2R)

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We will use a tree diagram to represent the
stages in the event. In order to keep the tree
manageable, we will define each stage as simply
as we can. i.e.success or not a success
The second stage is to choose the second
computer. Again there are two different
possibilities
The first stage has two possible outcomes
P(New New)
New
.5238
New
.2095
Refurb
P(New Refurb)
Choose a computer
.2095
P(Refurb New)
New
Refurb
Refurb
P(Refurb Refurb)
.0571
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Mean and Standard Deviation

• Now use these probabilities to create the
  probability distribution.

Amount of Loss 0 100 1000
Probability .5238 .4190 .0571
Note the probability of losing 100 happens on 2
branches of the tree
We want to know on average, how much of a loss
the company would have if this happened month
after month.
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Mean and Standard Deviation

• Now use these probabilities to create the
  probability distribution.
• To find the mean (expected value) of this
  distribution we can use the following
• ?(x) E(x) ?X P(X)
• The value of the variable times its probability
  for each, then find the sum.

Amount of Loss 0 100 1000
Probability .5238 .4190 .0571
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Mean and Standard Deviation

• So for our distribution
• ?(x) E(x) ?X P(X)
• M(x) E(x) 0(.5238) 100(.4190) 1000(.0571)
  99.00
• So the average loss for Knowway Computers is
  99.00

Amount of Loss 0 100 1000
Probability .5238 .4190 .0571
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Mean and Standard Deviation

• We can find the standard deviation in much the
  same way we find the standard deviation for a
  sample
• First, lets find the variance
• s2 ?(x-µ)2 P(x)
• (0 99)2 (.5238) (100-99)2(.4190)
  (1000-99)2(.0571) 51488.0199 Dollars2

Amount of Loss 0 100 1000
Probability .5238 .4190 .0571
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Mean and Standard Deviation

• So, the variance is 51488.0199 dollars2
• But we need the standard deviation which will
  bring our units back to dollars
• Remember that standard deviation is the square
  root of the variance so.
• s v51488.0199 266.91

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Mean and Standard Deviation

• We can use our calculators to find both the mean
  and standard deviation of a probability
  distribution.
• Enter the values of the RV in one list
• Enter the probabilities for each value into a
  second list
• Use Stat/Calc/1-var statistic (Xlist, Problist)
• The mean will be under x-bar
• The standard deviation will be sigma

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