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Probability Distributions. We need to develop probabilities of all possible distributions instead of just a particular/individual outcome Many probability.

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Presentation on theme: "Probability Distributions. We need to develop probabilities of all possible distributions instead of just a particular/individual outcome Many probability."— Presentation transcript:

1 Probability Distributions

2 We need to develop probabilities of all possible distributions instead of just a particular/individual outcome Many probability experiments have numerical outcomes which can be counted or measured A random variable X has a single value for each outcome in an experiment. Ex. If ‘X’ is the number rolled with a die, then X has a different value for each of the 6 possible outcomes. Random variables can be discrete or continuous.

3 Discrete  values that are separate from each other (number of possible values can be small) Continuous  have an infinite number of possible values in a continuous interval. This chapter involves discrete random variables. Classify each as either random or continuous. - number of customers on a paper route - Length of time it takes to deliver the papers - amount of money that you can make delivering papers Ask yourself if the values can be measured to an infinite point (ie. Does it require integers or fractions of a decimal.

4 Uniform Probability Distributions This just means that we need to calculate the probability of a group A,B,C,D or E being chosen for a particular place. Ex. Determine the probability distribution for comparing the probabilities of Group A presenting their project first, second, third, fourth or fifth.

5 Since each group has an equal probability for choosing each of the five positions, each prob is 1/5. Heading Random Variable, x Probability, P(x) Position 1 1/5 All outcomes are equally likely in a single trial. This distribution has a uniform probability distribution. The sum of the probabilities must equal 1 since it includes all possible outcomes.

6 For all values of x, P(X) = 1/n, where n is the number of possible outcomes in the experiment. Expected values E(x) is the predicted average of all possible outcomes of an experiment. The expectation is equal to the sum of the products of each outcome with its probability E(x) = x 1 P(x 1 ) + … + x n P(x n ) =

7 Consider a game in which you gain the points if it is an even number, and if you roll an odd number you lose that number of points. Show the distribution of points in this game using a table. Headings Number on Upper Face Points,x Probability, P(X) What is the expected outcome? Is this a fair game? Why?

8 It is not because the points gained and lost are not equal. For a game to be fair, the expected outcome must be 0. A building store has different sizes of 2x4s. There are 6- 6m boards, 5-8 m boards, 3-4m boards and 4-10m boards. Length of board (M), x Probability, P(x) 4m 3/18 6m 6/18 8m 5/18 10m 4/18

9 E(x) = 4(3/18) + 6(6/18) + 8(5/18) + 10(4/18) The expected length is 7.6444 Homework Pg 374 # 1,3,6,8,9


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