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Statistics Probability Distributions

Statistics Probability Distributions. Assignment 5 Example Problems. Discrete vs. Continuous. Discrete (Countable) Number of students in this class Number of points scored in a game Continuous (Measurable) Square footage of a house Time to complete a job. Probability Distributions.

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Statistics Probability Distributions

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  1. StatisticsProbability Distributions Assignment 5 Example Problems

  2. Discrete vs. Continuous • Discrete (Countable) • Number of students in this class • Number of points scored in a game • Continuous (Measurable) • Square footage of a house • Time to complete a job

  3. Probability Distributions • All probabilities • Must add to 1 • Must be positive • Must be between 0 and 1 • Where 0 is impossible and 1 is certain

  4. Mean • Finding the mean of a probability distribution

  5. Standard Deviation • Finding the standard deviation of a probability distribution This is the variance This is the standard deviation

  6. Probabilities • QUESTION: Multiple-choice questions each have five possible answers (a, b, c, d, e), one of which is correct. Assume that you guess the answers to three such questions. • Use the multiplication rule to find P(WWC), where C denotes a correct answer and W denotes a wrong answer. • ANSWER: Ok, notice there are 5 choices to answer (a,b,c,d,e) so you have a 1 in 5 chance of getting the right answer and 4 in 5 chances of getting the wrong answer. This means • The probability of getting the answer correct = P(C) = 1/5 • The probability of getting the answer wrong = P(W) = 4/5 • So if we get Wrong and Wrong and Correct this would be P(WWC) = (4/5)(4/5)(1/5) = 0.128 NOTE: The word “AND” in probabilities means to multiply

  7. Probabilities (continued) • QUESTION: Beginning with WWC, make a complete list of the different possible arrangements of one correct answer and two wrong answers and then find the probability for each entry in the list. • ANSWER: One correct and two wrong would be • WWC, WCW, CWW • P(WWC) = what we got previously= 0.128 • P(WCW) = (4/5)(1/5)(4/5) = see order does not matter with multiplication so = 0.128 • P(CWW) = (1/5)(4/5)(4/5) = see order does not matter with multiplication so = 0.128

  8. Probabilities (continued) • QUESTION: Based on the preceding results, what is the probability of getting exactly one correct answer when three guesses are made? • ANSWER: So this means P(WWC) OR P(WCW) OR P(CWW) 0.128 + 0.128 + 0.128 = 0.384 NOTE: The word “OR” in probabilities means to add

  9. Binomial Probabilities • QUESTION: Assume that a procedure yields a binomial distribution with a trial repeated n times. • Use the binomial probability of x successes given the probability p of success on a single trial. • n = 9, x = 6, p = 0.65 • Find P(6)

  10. Binomial Probabilities (calculator) • Calculator1) 2nd Vars2) Scroll down to binompdf3) The order is n, p, x so we would enter binompdf(9, .65, 6)    <--be sure to close the parentheses4) press enter to get 0.272 rounded

  11. Binomial Probabilities (by hand) • P(6) = = • now I am going to cancel  6*5*4*3*2*1 on top and bottom to be left with • cancel the 3 at the bottom with the 9 and cancel the 2 at the bottom with the 8 on top to getnow i put this in my calculator to get 0.272 rounded  WHEW!!! 

  12. Binomial Probabilities (2nd example) • QUESTION: A brand name has a 70% recognition rate. If the owner of the brand wants to verify that rate by beginning with a small sample of 10 randomly selected consumers, find the probability that exactly 7 of the consumers recognize the brand name. Also find the probability that the number who recognize the brand name is not 7.

  13. Binomial Probabilities (2nd example) • ANSWER:What we know • n = 10 (number in sample) • x = 7 (number of successes) • p = 70% or 0.70 • We want to find the probability of exactly 7 P(7) = in your TI83 press • 2nd Vars • Scroll down to binompdf • The order is n, p, x so we would enter binompdf(10,.7,7) • Press enter to get 0.267 (rounded) • Then “not 7” would be the complement of 1 minus the probability of 7 or P(not 7) = 1 – P(7) = 1 – 0.267 = 0.733

  14. Other examples • Other examples posted in same folder

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