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Transcript of Chapter 7 Random Variables I can find the probability of a discrete random variable. 6.1a Discrete...
Chapter 7Random Variables
I can find the probability of a discrete random variable.
6.1aDiscrete and Continuous Random
Variables and Expected Valueh.w: pg pg 353: 1 – 13 odd
Discrete and Continuous Random Variables
A random variable is a quantity whose value changes.
Discrete Random Variable
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 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 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
X = # heads X = 0, X = 1, X = 2, X = 3, X = 4
b. Find each probabilityP(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 of heads
0 1 2 3 4
Probability:
0.0625
c. Describe the probability histogram.
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?
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?
P(X ≥ 1): use the complement rule = 1 – P(X=0) = 1 – 0.0625
= 0.9375
Exercise: Roll of the Die
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?
P(X<3) = P(X=1) + P(X=2)
= 1/6 + 1/6 = 2/6 = 1/3
= 0.33
b. Use the TI-83/89 to:
Simulate rolling a die 100 times and assign the values to L1.
MATH:PRB:randInt(1,6,100) store L1
Sort the list in ascending order. STAT:SortA(L1)
Count the outcomes of 1 or 2 and record the relative frequency.
Results will vary but what do we expect the probability to be?
33% P(X=1) = /100 P(X=2) = /100 P(X=1) + P(X=2) =
Exercise: Three Children 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.
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.
BBB, BBG, BGB, GBB, GGB, GBG, BGG, GGG
What is the probability that X = 2? 3/8 = 0.375
c. Starting from your work in (a), find the distribution of X. That is, what values can X take, and what are
the probabilities of each value? (Hint: make a table.)
Value of X 0 1 2 3
Probability
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 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 VariableNow 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 You pick a 3 digit number. If your number is
chosen you win $500. There are 1000, 3 digit numbers. Each pick costs $1.
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?
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
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.
1 1 2 2X k k
i i
x p x p x p
x p
Example: Benford’s LawRecall: 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:
Find the μv:
=
V 1 2 3 4 5 6 7 8 9
Prb. .301 .176 .125 .097 .079 .067 .058 .051 .046
= 3.441
The means reflect the greater probability of smaller digits under Benfors’s law.
Histograms of μx and μv
We can’t locate the mean of a right skew distribution by eye – calculation is needed.