RVCND - Discrete Random Variables Lesson

Discrete random variables are shown using a probability model (fancy name for TABLE) showing X and P(X) as pictured here. Notice that the probabilities vary but still have a total sum = 1.

p of x discrete random var

The mean μ (mu) of the random variable is called the "expected value" and is sometime shown as E(X) where E(X) = SUM of the PRODUCT of all the X's and P(X)'s.

Using mathematical notation looks like ∑ (X) (P(X).

E(X) = (0)(.20) + (1)(.30) + (2)(.30) + (3)(.20)
For the above given probabilities of X values we would EXPECT the event to occur approximately 1.5 times (in theory) because our random variable values are integers.

The mean of a random variable is the average of the variable in two senses:

1) by definition it is the average of the possible values, weighted by their probabilities

2) by the law of large numbers it is the long run average of many independent observations on the variable.

The expected value is a weighted average since each probability is different. The procedure for finding the mean is similar to calculating your grade average where different categories are weighted differently like tests, homework, projects, etc. It is wise to use a calculator to assist with the math.
Enter the X values into List 1
Enter the P(X) values into List 2
DEFINE L3 as (L1)(L2)
Find the sum of L3 using the 2nd list command followed by MATH #5 sum.

Once we have the mean value we can continue to find the standard deviation. We also define the standard deviation (σ) called sigma by first defining the variance.

VARIANCE: σ ² = ∑ (X - μ ) ² (P(X))

STANDARD DEVIATION: σ = √ ( σ ² )

Continuing with the problem above, we substitute to get
Var = (0-1.5)² (.20 ) + (1 - 1.5)² (.30 ) + (2 - 1.5)² (.30) + (3 - 1.5)² (.20 )

Var = (2.25)(.20) + (.25)(.30) + (.25)(.30) + (2.2 5)(.20)

Var = 1.05 and σ = √ (1.05) = 1.02

Suppose 75% of all drivers always wear their seatbelts. Let's investigate how many of the drivers might be belted among five cars waiting at a traffic light. The possible outcomes range from no drivers wearing seatbelts through all five drivers wearing seatbelts. Using some ideas from the special probability unit we can determine the distribution associated with this question. Remember, the driver will either be wearing the seatbelt or not. Sounds binomial in nature!

Calculate the actual probability model.
MAKE A TABLE WITH THE PROBABILITIES FOR EACH RANDOM VARIABLE VALUE.

P(1) from binompdf(5, .75, 1) = .0146
P(2) from binompdf(5, .75, 2) = .0879
P(3) from binompdf(5, .75, 3) = .2637
P(4) from binompdf(5, .75, 4) = .3955
P(5) from binompdf(5, .75, 5) = .2373

Knowing 5 of the 6 probabilities will allow us to find the missing value.
All of the known values total .999 so P(0) = 1 - .999 or .001.

Using this theoretical probability model we would use the formula
E(X) = (0)(.001) + (1)(.0146) + (2)(.0879) + (3)(.2637) + (4)(.3955) + (5)(.2373).

It is best to use calculator lists to accomplish this. Pretty quickly we can determine that the expected number of seatbelt wearers given the specified conditions is approximately 3.75 or almost 4. Don't count on these numbers to be integers. The formula for standard deviation is a little more complex but the basic idea of substitution into the formula is the same.

More detail and examples are contained in the lessons below. Click on each lesson to download.

Discrete Random Variables Mean Expected Value

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