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Random Variables A random variable is a variable (usually we use x), that has a single numerical value, determined by chance, for each outcome of a procedure A probability distribution is a graph, table, or formula that gives the probability for each value of the random variable.

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Example of a probability distribution and probability histogram Outcome of the toss of two coins x (number of heads)P(x) 0 ¼ 1 ½ 2 ¼

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This is very similar to what we did in Chapter 2, except in Chapter 2 we were constructing frequency distributions and histograms based on observed data. In this chapter, we are doing similar things, but to describe what we expect to happen.

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Requirements for a Probability Distribution Where x takes on every possible value. Means: Something has to happen, so sum of all probabilities is 1. for each x. Means: the probability of any event is between 0 and 1 (inclusive)

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Center For a probability distribution, This gives the mean of the probability distribution. In other words, it tells us what the average (mean) outcome is. Notice how each outcome is weighted by how likely it is to occur. This is also called the Expected Value of the distribution, denoted E or E(x).

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Example Outcome of the toss of two coins (number of heads) xP(x)x. P(x) 0 ¼ 0. ¼ = 0 1 ½ 1. ½ = ½ 2 ¼ 2. ¼ = ½

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Example Number of matching digits (of a guess to a random 2-digit number) xP(x)x. P(x) 0 0.81 0(0.81) = 0 1 0.18 1(0.18) = 0.18 2 0.01 2(0.01) = 0.02 So on average, we will match 0.2 digits. So over 1000 tries, we would expect to match about 200 digits total.

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Variance For a probability distribution This gives the variance of the probability distribution. In other words, it tells us how spread out we expect the outcomes to be. As before, Standard Deviation can be found from the variance:

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Example Outcome of the toss of two coins (number of heads) xP(x)x. P(x)x 2. P(x) 0 ¼ 0. ¼ = 00 2. ¼ = 0 1 ½ 1. ½ = ½1 2. ½ = ½ 2 ¼ 2. ¼ = ½2 2. ¼ = 1

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Example Number of matching digits (of a guess to a random 2-digit number) xP(x)x. P(x)x 2. P(x) 0 0.81 0(0.81) = 00 2 (0.81) = 0 1 0.18 1(0.18) = 0.181 2 (0.18) = 0.18 2 0.01 2(0.01) = 0.022 2 (0.01) = 0.04

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You try Number of Girls out of 4 children xP(x) 01/16 = 0.0625 1¼ = 0.25 23/8 = 0.375 3¼ = 0.25 41/16 = 0.0625

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Homework Part 1 4.2: 3, 5, 7 Read the rest of section 4.2, especially “Identifying Unusual Results with Probabilities”, pg 189.

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Excursions in Modern Mathematics, 7e: 16.5 - 1Copyright © 2010 Pearson Education, Inc. 16 Mathematics of Managing Risks Weighted Average Expected Value.

Excursions in Modern Mathematics, 7e: 16.5 - 1Copyright © 2010 Pearson Education, Inc. 16 Mathematics of Managing Risks Weighted Average Expected Value.

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