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.