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Random Variables & Probability Distributions

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11 diapositives · 329 Views · Jul 20, 2014

Random Variables & Probability Distributions. “The probability of someone laughing at you is proportional to the stupidity of your actions.”. Definitions Random Variable: A variable whose possible values are numerical outcomes of a random phenomenon Discrete Random Variable:

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  1. Random Variables & Probability Distributions “The probability of someone laughing at you is proportional to the stupidity of your actions.”

  2. Definitions Random Variable: A variable whose possible values are numerical outcomes of a random phenomenon Discrete Random Variable: Has a countable number of outcomes; distribution described by a histogram Continuous Random Variable: Takes all values in an INTERVAL of numbers; distribution described by a density curve (i.e., normal curve)

  3. Definitions Sample Space: Set of all possible outcomes Probability Distribution: A chart that lists the values of the discrete random variable X and the probability that X occurs; all probabilities add up to 1; can graph as a histogram

  4. Expected Value of a Discrete Random Variable Expected Value = mean = *this is the sum of the products of x and P(x) from the probability distribution

  5. Example 1: State whether each of the following random variables is discrete or continuous: • a) The number of defective tires on a car. • discrete • b) The body temperature of a hospital patient. • continuous • c) The number of pages in a book. • discrete • d) The lifetime of a light bulb. • continuous

  6. Example 2: Let X denote the number of broken eggs in a randomly selected carton of one dozen “store brand” eggs at a certain market. Suppose that the probability distribution of X is as follows: • a) Interpret P(X = 1) =.20. • The probability that there is 1 broken egg is .20. • b) Find the probability that there are exactly 4 broken eggs. • P(X = 4) = .01 • c) Find the probability that there are at least 2 broken eggs. • P(X = 2 or X = 3 or X = 4) = .10 + .04 + .01 = .15 • d) Find the probability that there are less than 2 broken eggs. • P(X = 0 or X = 1) = .65 + .20 = .85 • e) How many eggs should we “expect” to be broken? • E(X) = 0(.65) + 1(.20) + 2(.10) + 3(.04) + 4(.01) =

  7. Example 3. Suppose we flip a coin 3 times. Let X = the number of heads. • Write the sample space of flipping the coin. • HHH • HHT • HTH • THHHTT • THT • TTH • TTT

  8. Example 3. b) Determine the probability distribution.

  9. Example 3. • c) Sketch a histogram of the probability distribution.

  10. Example 3. • d) How many heads would we “expect”? • E(X) =

  11. Example 4: Suppose we toss two dice and sum the results. a) Write out the sample space b) Determine the probability distribution. c) What is the probability you get a sum less than 5? d) What is the probability you get a sum of at least 10? e) What is the expected value of the sum of the two dice?