Transcript 6_1a Random Variables and Expected Value

Chapter 6
Random Variables
I can find the probability of a discrete random
variable.
6.1a
Discrete and Continuous Random
Variables and Expected Value
Discrete and Continuous Random
Variables
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A random variable is a quantity whose
value changes.
Discrete Random Variable
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A discrete random variable is a variable
whose value is obtained by counting.
number of students present
number of red marbles in a jar
number of heads when flipping three coins
students’ grade level
Probability Distribution
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The probability distribution of X lists the
values and their probabilities.
Value of X: x1, x2, x3, … , xk
Probability: p1, p2 , p3 , … , pk
To find the probability of event pi , add up
the probabilities of the xi that make up
that event.
Example: Getting Good Grades
A teacher gives the following grades:
 15% A’s and D’s, 30% B’s, C’s; 10% F’s
 on a 4 point scale (A=4).
 Chose a student at random and find the
probability they get a B or better.
Here is the distribution of X:
Grade: 0
1
2
3
4
Prob: .10 .15 .30 .30 .15
 P(get a B or better) s/a P(grade 3 or 4):
 P(X=3) + P(X=4)
 = 0.30 + 0.15
 = 0.45
Probability Histograms
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Height of each bar is
the probability
Heights add up to 1
Prob. of Benfords Law
vs. random digits
Example: Tossing Coins
a. Find the probability distribution of the
discrete random variable X that counts the
# heads in 4 tosses of a coin.
 Assume: fair coin, independence
 Determine the # of possible outcomes
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X = # heads
X = 0, X = 1, X = 2, X = 3, X = 4
b. Find each probability
P(X=0) = 1/16 = 0.0625
P(X=1) =
P(X=2) =
P(X=3) =
P(X=4) =
 Do they add up to 1?
 Yes, so legitimate
distribution.
Make a table of the probability
distribution.
Number 0
of
heads
Probabil 0.0625
ity:
1
2
3
4
c. Describe the probability histogram.
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It is exactly symmetric.
It is the idealization of the relative frequency
distribution of the number of heads after many
tosses of four coins.
d. What is the prob. of tossing at least 2 heads?
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P(X ≥ 2) = P(X=2) + P(X=3) + P(X=4)
= 0.375 + 0.25 + 0.0625
= 0.6875
e. What is the prob. of tossing at least 1 head?
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P(X ≥ 1): use the complement rule
= 1 – P(X=0)
= 1 – 0.0625
= 0.9375
Exercise: Roll of the Die
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If a carefully made die is rolled once, is it
reasonable to assign probability 1/6 to each
of the six faces?
a. What is the probability of rolling a
number less than 3?
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P(X<3) = P(X=1) + P(X=2)
= 1/6 + 1/6 = 2/6 = 1/3
= 0.33
Exercise: Three Children
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A couple plans to have three children.
There are 8 possible arrangements of girls
and boys.
For example, GGB. All 8 arrangements are
approximately equally likely.
a. Write down all 8 arrangements of
the sexes of three children.
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What is the probability of any one of these
arrangements?
BBB, BBG, BGB, GBB, GGB, GBG, BGG,
GGG
Each has probability of 1/8
b. Let X be the number of girls the
couple has.
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BBB, BBG, BGB, GBB, GGB, GBG, BGG,
GGG
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What is the probability that X = 2?
3/8 = 0.375
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c. Starting from your work in (a), find
the distribution of X.
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That is, what values can X take, and what are
the probabilities of each value?
(Hint: make a table.)
Value of X
Probability
0
1
2
3
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X is the number of girls the couple has.
Value of X
0
1
2
3
Probability
1/8
3/8
3/8
1/8
Review of Probability
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The probability of a random variable is an
idealized relative frequency distribution.
Histograms and density curves are
pictures of the distributions of data.
When describing data, we moved from
graphs to numerical summaries such as
means and standard deviations.
The Mean of a Random Variable
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Now we will make the same move to
expand our description of the distribution of
random variables.
The mean of a discrete random variable, X,
is its weighted average.
Each value of X is weighted by its
probability. Not all outcomes need to be
equally likely.
Example: Tri-Sate Pick 3
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You pick a 3 digit number. If your number is
chosen you win $500. There are 1000, 3
digit numbers. Each pick costs $1.
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Taking X to be the amount your ticket pays
you, the probability distribution is:
Payoff X:
$0
$500
Probability: 0.999 0.001
Find your average Payoff.
Normal “average”:
(0 + 500) /2 =$250
Are the outcomes equally likely?
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The long run weighted average is:
= 500(1/1000) + 0(999/1000)
= $ 0.50
Conclusion:
In the long run, the state keeps ½ of what
you wager.
Expected Value (μx)
(The long run average outcome)
 We do not expect one observation to
be close to its expected value.
 μx : probabilities add to 1
Mean of a Discrete Random
Variable
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The mean of a discrete random variable, X,
is its weighted average. Each value of X is
weighted by its probability.
To find the mean of X, multiply each value
of X by its probability, then add all the
products.
 X  x1 p1  x2 p2    xk pk
  xi pi
Example: Benford’s Law
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Recall: If the digits in a set of data appear
“at random,” the nine possible digits 1 to 9
all have the same probability each being
1/9.
The mean of the distribution is:
μx = 1(1/9) + 2(1/9) + … + 9(1/9)
= 45(1/9)
=5
But, if the data obey Benford’s law,
the distribution of the first digit V is:
V
1
2
3
4
5
6
7
8
9
Prb. .301 .176 .125 .097 .079 .067 .058 .051 .046
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Find the μv:
=
= 3.441
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The means reflect the greater probability of
smaller digits under Benfors’s law.
Histograms of μx and μv
can’t locate the mean of a right
skew distribution by eye – calculation
is needed.
 We