Chapter-6_Note

# Chapter-6_Note - 6 Probability Distributions 6.1...

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 6 Probability Distributions 6.1 Summarizing Possible Outcomes and Their Probabilities Random Variable A random variable is a numerical measurement of the outcome of a random phenomenon. Often, the randomness results from the use of random sampling or a randomized experiment to gather the data. Ex.1: Number of touchdowns for Red Raiders, Duration of a call Probability Distribution The probability distribution of a random variable specifies its possible values and their probabilities. Ex.2: Probability distribution of number of home runs in a game for Boston Red Sox. The table lists the possible values for the number of home runs and the corresponding probabilities. Number of Home Runs Probability 0.23 1 0.38 2 0.22 3 0.13 4 0.03 5 0.01 6 or more 0.00 Probability distributions can be defined for both discrete random variables and continuous random variables. We will first look at discrete random variables. Probability Distribution of a Discrete Random Variable A discrete random variable X takes a set of seperate values (such as 0,1,2,....). Its probability distribution assigns a probability P ( x ) to each possible value of x . • For each x , the probability P ( x ) falls between 0 and 1. • The sum of the probabilities for all the possible x values equals 1. 1 Ex.3: Refer to Ex.2 . • for each x , the probability falls between 0 and 1. (all the values 0.23, 0.38, 0.22, 0.13, 0.03, and 0.01 are between 0 and 1) • sum of the probabilities for all the possible x values equals 1. (0.23 + 0.38 + 0.22 + 0.13 + 0.03 + 0.01 = 1) Therefore, it is a probability distribution. Probability of at least 3 homeruns is P ( X ≥ 3) = P (3) + P (4) + P (5) + P (6) = 0 . 13 + 0 . 03 + 0 . 01 + 0 . 00 = 0 . 17 Figure 1: Probability Distribution for Ex.2 Ex.4: From six marbles numbered as 1,1,1,1,2,2, two marbles will be drawn at ran- dom without replacement. Let X denote the sume of the numbers on the selected marbles. List the possible values of X and determine the probability distribution. 2 X- sum of the two numbers The probability distribution for this experiment is given in the following table. X Probability 2 6 15 3 8 15 4 1 15 The Mean of a Probability Distribution ( μ ) The mean of a probability distribution for a discrete random variable is μ = summationdisplay xP ( x ) where the sum is taken over all possible values of x . This is also called the expected value for X . Ex.5: The mean of the number of homeruns for in a game for Red Sox is, μ = 0(0 . 23) + 1(0 . 38) + 2(0 . 22) + 3(0 . 13) + 4(0 . 03) + 5(0 . 01) = 1 . 38 This also means that you can expect the Red Sox to have an average of 1.38 homeruns per game for that season. Probability Distribution of a Continuous Random Variable A continuous random variable has possible values that form an interval. Its probability distribution is specified by a curve that determines the probability that the random variable falls in any particular interval of values....
View Full Document

{[ snackBarMessage ]}

### Page1 / 10

Chapter-6_Note - 6 Probability Distributions 6.1...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online