1. The Practice of Statistics Third Edition Chapter 7: Random Variables 7.2 The mean and Standard Deviation of a Random Variable Copyright © 2008 by W. H. Freeman & Company Daniel S. Yates

2. Some of the examples and slides were copied from the following source: Roxy Peck Chris Olsen Jay Devore Introduction to Statistics & Data Analysis 3ed Thompson Brooks/Cole, a part of The Thompson Corporation

3. Essential Questions How do you calculate the mean of a discrete random variable? How do you calculate the variance and standard deviation of a discrete random variable? What is the law of large numbers? Given µ X and µ Y, How do you calculate µ a+bX and µ X+Y ? Given X and Y are independent, how do you calculate σ 2 a+bX and σ 2 X+Y ? How do you describe the shape of a linear combination of independent Normal random variables?

4. Review

7. Calculating The Means of a Discrete Random Variable Distribution

8. Example 1 A professor regularly gives multiple choice quizzes with 5 questions. Over time, he has found the distribution of the number of wrong answers on his quizzes is as follows

9. Example Continued Multiply each x value by its probability and add the results to get  x.  x = 1.41

10. Example – Apgar Scores At I min after birth and again at 5 min, each newborn child is given a numerical rating called the Apgar score. Possible values of the score are 0, 1, 2, …, 9, 10. A child’s score is determined by five factors: muscle tone, skin color, respiratory effort, strength of heartbeat, and reflex. A high score indicates a healthy baby. Let X denote the Apgar score at I min of a randomly selected newborn infant at a particular hospital. The probability distribution is: 012345678910.002.001.002.005.02.04.17.38.25.12.01 X P(X) What is the mean Apgar score at this hospital? µ X = 7.16

11. Calculating Variance

12. Example 1 Again A professor regularly gives multiple choice quizzes with 5 questions. Over time, he has found the distribution of the number of wrong answers on his quizzes is as follows µ X = 1.41 Calculate the variance for the of X.

13. Example - continued

14. Example – Apgar Score Again 012345678910.002.001.002.005.02.04.17.38.25.12.01 X P(X) µ X = 7.16 Find the variance and standard deviation for the discrete random variable X.

15. Number of Observations versus Accurate Estimates of μ

17. Rules for Means and Variances

18. Suppose Suppose we want to find the means length and Standard Deviation of a species of grasshoppers in a particular field. We run our survey and establish the following distribution: X = the length in inches X123 P(X)0.20.50.3 Find the mean and standard deviation of X. μ X = 2.1 inches σ 2 X = 0.49 σ = 0.7 inches

19. More Supposing Suppose we are required to report our findings using the metric system. How does changing to centimeters affect µ X and σ X ? X (in.)123 P(X)0.20.50.3 2.54X2.54(1)2.54(2)2.54(3) P(X)0.20.50.3 μ X = 2.1 in. σ 2 X = 0.49 µ 2.54X = 2.54(1)(0.2) + 2.54(2)(0.5) + 2.54(3)(0.3) = 2.54[(1)(0.2) + (2)(0.5) + (3)(0.3)] = 2.54[ 2.01] = 2.54µ x --------- Let b = 2.54, then bµ x σ 2 2.54X = ( 2.54(1) – 2.54(2.1)) 2 (0.2) + (2.54(2) – 2.54(2.1)) 2 (0.5) + (2.54(3) – 2.54(2.1)) 2 (0.3) = (2.54(-1.1)) 2 (0.2) + (2.54(-0.1)) 2 (0.5) + (2.54(0.9)) 2 (0.3) = 2.54 2 ((-1.1) 2 (0.2) + (-0.1) 2 (0.5) + (0.9) 2 (0.3)) = 2.54 2 (0.49) = 2.54 2 σ 2 X --------------- b 2 σ 2 X

20. Even More Supposing We discover after the survey that the instruments we used to measure the grasshopper were miss calibrated. The measurements were 0.1 inches too short. Is there a way to adjust the results without re-doing the entire survey? 0.1 inches =.254 cm.254 + 2.54X.254 + 2.54.254 + 5.08.254 + 7.62 P(X)0.20.50.3 µ.254+2.54X = (.254 + 2.54)(0.2) + (.254 + 5.08)(0.5) + (.254 + 7.62)(0.3) =.254(.2) +.254(.5) +.254(.3) +2.54(.2) + 5.08(.5) +(7.62)(.3) = [.254(.2) +.254(.5) +.254(.3)] +[2.54(.2) + 5.08(.5) +(7.62)(.3)] =.254[ 1(.2) + 1(.5) + 1(.3)] + 2.54[ 1(.2) + 2(.5) + 3(.3)] =.254 + 2.54µ X If we let a =.254 and b = 2.54, then we have µ a+bX = a + bµ x

21. Rules for Means

23. Example 2 Suppose x is the number of sales staff needed on a given day. If the cost of doing business on a day involves fixed costs of $255 and the cost per sales person per day is $110, find the mean cost (the mean of x or  x ) of doing business on a given day where the distribution of x is given below.

24. Example 2 continued We need to find the mean of y = 255 + 110x

25. Example 2 continued We need to find the variance and standard deviation of y = 255 + 110x

26. 1.  y = a 1  1 + a 2  2 +  + a n  n (This is true for any random variables with no conditions.) Means and Variances for Linear Combinations 2.If x 1, x 2, , x n are independent random variables then and If x 1, x 2, , x n are random variables with means  1,  2, ,  n and variances respectively, and y = a 1 x 1 + a 2 x 2 +  + a n x n then

27. Example 3 A distributor of fruit baskets is going to put 4 apples, 6 oranges and 2 bunches of grapes in his small gift basket. The weights, in ounces, of these items are the random variables x 1, x 2 and x 3 respectively with means and standard deviations as given in the following table. Find the mean, variance and standard deviation of the random variable y = weight of fruit in a small gift basket. ApplesOrangesGrapes Mean  8107 Standard deviation  0.91.12

28. Example 3 continued It is reasonable in this case to assume that the weights of the different types of fruit are independent. ApplesOranges Mean  8107 Standard deviation  0.91.12 Grapes

29. Example 4 A nationwide standardized exam consists of a multiple choice and a free response section. Each section, the mean and standard deviation are reported to be: Let X 1 = the score for multiple-choice and X 2 = the score for the free response. Suppose the total score is computed as Y = X 1 + 2X 2 Find the mean and standard deviation for the test MeanStandard Deviation Multiple Choice386 Free Response307

30. Example 4 Continued The Mean The Standard Deviation Are X 1 and X 2 independent? It is likely that X 1 and X 2 are not independent since a person who does well on multiple choice section will also do well on the free response section and vice versa. Therefore, we can not calculate the standard deviation with the given information.

31. Last Problem Suppose that the mean height of policemen is 70 inches with a standard deviation of 3 inches. And suppose that the mean height for policewomen is 65 inches with a standard deviation of 2.5 inches. If heights of policemen and policewomen are Normally distributed, find the probability that a randomly selected policewoman is taller than a randomly selected policeman.

32. Last Problem - Continued Let X = height of policewoman and Y= height of policeman. μ X-Y = 65 – 70 = -5. σ 2 X-Y = σ 2 X + σ 2 Y = 2.5 2 + 3 2 = 15.25 σ X-Y = Using N(-5, 3.905) find P(X –Y < 0). From the Calculator ( NormCDF ( -9999, 0, -5, 3.905) P(X – Y < 0) = 0.8997

33. Summary The Law of Large Numbers says that the average of the values of X observed in many trials approach μ. If X is discrete with the possible values x i, the means is the average of the values of X, each weighted by its probability: μ X = x 1 p 1 + x 2 p 2 + … + x n p n The variance σ 2 for a discrete variable: σ 2 X = (x 1 - μ) 2 p 1 + (x 2 – μ) 2 p 2 + … + (x n - μ) 2 p n The standard deviation σ X is the square root of the variance.

34. Summary The means and variance of random variables obey the following rules If a and b are fixed numbers, then μ a + bX = a + b μ X σ 2 a+bX = b 2 σ 2 X If X and Y are any two random variables, then μ X + Y = μ X + μ Y And if X and Y are independent, then σ 2 X + Y = σ 2 X + σ 2 Y σ 2 X – Y = σ 2 X + σ 2 Y