1. Copyright © Cengage Learning. All rights reserved. 8 PROBABILITY DISTRIBUTIONS AND STATISTICS

2. Copyright © Cengage Learning. All rights reserved. 8.3 Variance and Standard Deviation

3. 3 Standard Deviation

4. 4 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.

5. 5 Standard Deviation

6. 6 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.

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

8. 8 Applied Example 3 – Solution 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 cont’d

9. 9 Applied Example 3 – Solution 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 cont’d

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

11. 11 Applied Example 3 – Solution 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. cont’d

12. 12 Chebychev’s Inequality

13. 13 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.

14. 14 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 Figure 10 At least 75% of the outcomes fall within this interval.

15. 15 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). Figure 11 At least 89% of the outcomes fall within this interval.

16. 16 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.

17. 17 Example 4 – Solution 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%. cont’d

18. 18 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.

19. 19 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.

20. 20 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 – 6k = 30 and 40 + 6k = 50 from which we deduce that

21. 21 Applied Example 5 – Solution 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%. cont’d

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

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