8 PROBABILITY DISTRIBUTIONS AND STATISTICS Copyright Cengage Learning

8. 3 Variance and Standard Deviation Copyright © Cengage Learning. All rights reserved.

Standard Deviation 3

Standard Deviation Because Equation (5), which gives the variance of the random variable X, involves the squares of the deviations, the unit of measurement of Var(X) is the square of the unit of measurement of the values of X. For example, if the values assumed by the random variable X are measured in units of a gram, then Var(X) will be measured in units involving the square of a gram. To remedy this situation, one normally works with the square root of Var(X) rather than Var(X) itself. The former is called the standard deviation of X. 4

Standard Deviation 5

Applied Example 3 – Packaging Let X and Y denote the random variables whose values are the weights of the Brand A and Brand B potato chips, respectively Compute the means and standard deviations of X and Y and interpret your results. 6

Applied Example 3 – Solution The probability distributions of X and Y may be computed from the given data as follows: 7

Applied Example 3 – Solution cont’d The means of X and Y are given by x = (. 1)(15. 8) + (. 2)(15. 9) + (. 4)(16. 0) + (. 2)(16. 1) + (. 1)(16. 2) = 16 y = (. 2)(15. 7) + (. 1)(15. 8) + (. 1)(15. 9) + (. 1)(16. 0) + (. 2)(16. 1) + (. 2)(16. 2) + (. 1)(16. 3) = 16 8

Applied Example 3 – Solution cont’d Therefore, Var(X) = (. 1)(15. 8 – 16)2 + (. 2)(15. 9 – 16)2 + (. 4)(16 – 16)2 + (. 2)(16. 1 – 16)2 + (. 1)(16. 2 – 16)2 = 0. 012 Var(Y) = (. 2)(15. 7 – 16)2 + (. 1)(15. 8 – 16)2 + (. 1)(15. 9 – 16)2 + (. 1)(16 – 16)2 + (. 2)(16. 1 – 16)2 + (. 2)(16. 2 – 16)2 + (. 1)(16. 3 – 16)2 = 0. 042 9

Applied Example 3 – Solution cont’d So the standard deviations are 10

Applied Example 3 – Solution cont’d The mean of X and that of Y are both equal to 16. Therefore, the average weight of a package of potato chips of either brand is 16 ounces. However, the standard deviation of Y is greater than that of X. This tells us that the weights of the packages of Brand B potato chips are more widely dispersed about the common mean of 16 than are those of Brand A. 11

Chebychev’s Inequality 12

Chebychev’s Inequality The standard deviation of a random variable X may be used in statistical estimations. For example, the following result, derived by the Russian mathematician P. L. Chebychev (1821– 1894), gives a bound on the proportion of the values of X lying within k standard deviations of the expected value of X. 13

Chebychev’s Inequality To shed some light on this result, let’s take k = 2 in Inequality (9) and compute P( – 2 X + 2 ) 1 – =1– =. 75 This tells us that at least 75% of the outcomes of the experiment lie within 2 standard deviations of the mean (Figure 10). Taking k = 3 in Inequality (9), we have At least 75% of the outcomes fall within this interval. Figure 10 14

Chebychev’s Inequality P( – 3 X + 3 ) 1 – =1– = . 89 This tells us that at least 89% of the outcomes of the experiment lie within 3 standard deviations of the mean (Figure 11). At least 89% of the outcomes fall within this interval. Figure 11 15

Example 4 A probability distribution has a mean of 10 and a standard deviation of 1. 5. Use Chebychev’s inequality to find a bound on the probability that an outcome of the experiment lies between 7 and 13. Solution: Here, = 10 and = 1. 5. To determine the value of k, note that – k = 7 and + k = 13. Substituting the appropriate values for and , we find k = 2. 16

Example 4 – Solution cont’d Using Chebychev’s Inequality (9), we see that a bound on the probability that an outcome of the experiment lies between 7 and 13 is given by —that is, at least 75%. 17

Chebychev’s Inequality Note: The results of Example 4 tell us that at least 75% of the outcomes of the experiment lie between 10 – 2 and 10 + 2 —that is, between 7 and 13. 18

Applied Example 5 – Industrial Accidents Great Northwest Lumber Company employs 400 workers in its mills. It has been estimated that X, the random variable measuring the number of mill workers who have industrial accidents during a 1 -year period, is distributed with a mean of 40 and a standard deviation of 6. Using Chebychev’s Inequality (9), find a bound on the probability that the number of workers who will have an industrial accident over a 1 -year period is between 30 and 50, inclusive. 19

Applied Example 5 – Solution Here, = 40 and = 6. We wish to estimate P(30 X 50). To use Chebychev’s Inequality (9), we first determine the value of k from the equation – k = 30 or + k = 50 Since = 40 and = 6 in this case, we see that k satisfies 40 – 6 k = 30 and 40 + 6 k = 50 from which we deduce that 20

Applied Example 5 – Solution cont’d Thus, a bound on the probability that the number of mill workers who will have an industrial accident during a 1 -year period is between 30 and 50 is given by P(30 X 50) 1 – —that is, at least 64%. 21

Practice p. 462 #37 -40 p. 464 Technology Exercises #5 22

Assignment p. 458 Exercises #1 -6 Standard Deviation, 7, 9, 12, 14, 20, 24, 25, 32, 41 -44 23