StatisticsProbability Distributions

Assignment 5Example 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

• All probabilities– Must add to 1– Must be positive– Must be between 0 and 1• Where 0 is impossible and 1 is certain

Mean

• Finding the mean of a probability distribution

x P(x)

0 0.08631 0.49352 0.26983 0.10944 0.03195 0.0091

5344.10091.0*50319.0*41094.0*32698.0*24935.0*10863.0*0)(* xPx

Standard Deviation

• Finding the standard deviation of a probability distribution

x P(x)

0 0.08631 0.49352 0.26983 0.10944 0.03195 0.0091

2952.30091.0*50319.0*41094.0*32698.0*24935.0*10863.0*0)(* 2222222 xPx

940817.05344.12952.3)(* 222 xPx This is the variance

0.19699.0940817.0var iance This is the standard deviation

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

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

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

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)

Binomial Probabilities (calculator)

• Calculator

1) 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

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 get

now i put this in my calculator to get 0.272 rounded WHEW!!!

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.

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 7P(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

Other examples

• Other examples posted in same folder