CHAPTER 6 Random Variables 6.2 Transforming and Combining Random Variables The Practice of Statistics, 5th Edition Starnes, Tabor, Yates, Moore Bedford Freeman Worth Publishers
6.2 Reading Quiz (T or F) 1. When transforming a random variable, the mean is affected by adding, subtracting, multiplying and dividing. 2. When transforming a random variable, the standard deviation is affected by adding, subtracting, multiplying and dividing. 3. When combining two random variables, μ X+Y = μ X + μ Y. 2 = 4. When combining two independent random variables, σ X Y σ 2 2 X σ Y 5. The sum or difference of independent Normal random variables follows a Normal distribution. The Practice of Statistics, 5 th Edition 2
HW 22. Let Y be a number between 0 and 1 produced by a random number generator. Assuming that the random variable Y has a uniform distribution, find the following probabilities: a) P(Y 0.4) b) P Y < 0.4 c) P(0.1 < Y 0.15 or 0.77 Y < 0.88) (a) P Y 0.4 = 0.4 (b) P Y < 0.4 = 0.4 (c) P(0.1 < Y 0.15 or 0.77 Y < 0.88 = P 0.1 < Y 0.15 + P 0.77 Y < 0.88 = 0.15 0.1 + 0.88 0.77 = 0.05 + 0.11 = 0.16 The Practice of Statistics, 5 th Edition 3
HW 24. The Normal distribution with mean μ = 6.8 and standard deviation σ = 1.6 is a good description of the Iowa Test of Basic Skills (ITBS) vocabulary scores of seventh-grade students in Gary, Indiana. Call the score of a randomly chosen student X for short. Find P(X 9) and interpret the result. Step 1: State the distribution and values of interest. The score X of a randomly chosen student has the N(6.8, 1.6) distribution. We want to find P(X 9). Step 2: Perform calculations. Show your work. The standardized score for the boundary value is z = 9 6.8 1.6 = 1.38. = From Table A, the desired probability is P(Z 1.38) = 1-0.9162 = 0.0838. Using technology: The command normalcdf(lower: 9, upper: 1000, µ: 6.8, σ: 1.6) gives an area of 0.0846. Step 3: Answer the question. There is a 0.0846 probability of selecting a student with a score of at least 9. The Practice of Statistics, 5 th Edition 4
HW 26. The length of human pregnancies from conception to birth follows a Normal distribution with mean 266 days and standard deviation 16 days. Choose a pregnant woman at random. Let X = the length of her pregnancy. a) Find P(X 240) and interpret the result. (a) Step 1: State the distribution and values of interest. The length X of a randomly chosen woman s pregnancy has the N(266, 16) distribution. We want to find P(X 240). Step 2: Perform calculations. Show your work. The standardized score for the boundary value is z + 240 266 16 = 1.63. From Table A, the desired probability is P(Z 1.63) = 1 0.0516 = 0.9484. Using technology: The command normalcdf(lower: 240, upper: 1000, µ: 266, σ: 16) gives an area of 0.9479. Step 3: Answer the question. There is a 0.9479 probability of selecting a woman whose pregnancy lasts at least 240 days. b) What is P X > 240? Explain. The line above 240 has no area, so P(X 240) = P(X > 240) = 0.9479. The Practice of Statistics, 5 th Edition 5
HW 26. The length of human pregnancies from conception to birth follows a Normal distribution with mean 266 days and standard deviation 16 days. Choose a pregnant woman at random. Let X = the length of her pregnancy. c) Find the value of c such that P X c = 0.20. Show your work. Step 1: State the distribution and values of interest. The length X of a randomly chosen woman s pregnancy has the N(266, 16) distribution. We want to c such that P(X c) = 0.20 (see graph below). Step 2: Perform calculations. Look in the body of Table A for a value closest to 0.80. A z-score of 0.84 gives the closest value (0.7995). Solving 0.84 = c 266 16 gives c = 279.44. Using technology: The command invnorm(area: 0.80, µ: 266, σ: 16) gives a value of 279.47. Step 3: Answer the question. 20% of pregnancies last at least 279.47 days. The Practice of Statistics, 5 th Edition 6
6.1 Review: Size of American Households In government data, a household consists of all occupants of a dwelling unit, while a family consists of two or more persons who live together and are related by blood or marriage. So all families form households, but some households are not families. Here are the distributions of household size and family size in the United States. Number of persons 1 2 3 4 5 6 7 Household probability 0.25 0.32 0.17 0.15 0.07 0.03 0.01 Family Probability 0 0.42 0.23 0.21 0.09 0.03 0.02 Let X = the number of people in a randomly selected US household and Y = the number of people in a randomly chosen US family. a) Make histograms suitable for comparing the probability distributions of X and Y. Describe any differences that you observe. b) Find the mean for each random variable. Explain why this difference makes sense. c) Find and interpret the standard deviations of both X and Y. The Practice of Statistics, 5 th Edition 7
6.1 Review: Size of American Households a) Both distributions are skewed to the right. The center for the family distribution is greater than the center for the household distribution, but the spread of the family distribution is less that the spread of the household distribution. Also, the event X = 1 has a much higher probability in the household distribution. b) μ X = 1 0.25 + 2(0.32) + + 7 0.01 = 2.6 people. μ Y = 1 0 + 2(0.42) + + 7 0.02 = 3.14 people. The mean for families is larger, which agrees with observation in part a. c) σ X = 1 2.6 2 0.25 + 2 2.6 2 0.32 + + 7 2.6 2 (0.01) = 1.421 The number of people in a randomly selected household will typically differ from the mean of 2.6 by about 1.421 people. σ Y = 1 3.14 2 0 + 2 3.14 2 0.42 + + 7 3.14 2 (0.02) = 1.249 The number of people in a randomly selected family will typically differ from the mean of 3.14 by about 1.249 people. The Practice of Statistics, 5 th Edition 8
Transforming and Combining Random Variables Learning Objectives After this section, you should be able to: DESCRIBE the effects of transforming a random variable by adding or subtracting a constant and multiplying or dividing by a constant. FIND the mean and standard deviation of the sum or difference of independent random variables. FIND probabilities involving the sum or difference of independent Normal random variables. The Practice of Statistics, 5 th Edition 9
Linear Transformations In Section 6.1, we learned that the mean and standard deviation give us important information about a random variable. In this section, we ll learn how the mean and standard deviation are affected by transformations on random variables. In Chapter 2, we studied the effects of linear transformations on the shape, center, and spread of a distribution of data. Recall: 1. Adding (or subtracting) a constant, a, to each observation: Adds a to measures of center and location. Does not change the shape or measures of spread. 2. Multiplying (or dividing) each observation by a constant, b: Multiplies (divides) measures of center and location by b. Multiplies (divides) measures of spread by b. Does not change the shape of the distribution. The Practice of Statistics, 5 th Edition 10
Multiplying a Random Variable by a Constant Pete s Jeep Tours offers a popular half-day trip in a tourist area. There must be at least 2 passengers for the trip to run, and the vehicle will hold up to 6 passengers. Define X as the number of passengers on a randomly selected day. Passengers x i 2 3 4 5 6 Probability p i 0.15 0.25 0.35 0.20 0.05 The mean of X is 3.75 and the standard deviation is 1.090. Pete charges $150 per passenger. The random variable C describes the amount Pete collects on a randomly selected day. Collected c i 300 450 600 750 900 Probability p i 0.15 0.25 0.35 0.20 0.05 The mean of C is $562.50 and the standard deviation is $163.50. Compare the shape, center, and spread of the two probability distributions. The Practice of Statistics, 5 th Edition 11
Multiplying a Random Variable by a Constant Effect on a Random Variable of Multiplying (Dividing) by a Constant Multiplying (or dividing) each value of a random variable by a number b: Multiplies (divides) measures of center and location (mean, median, quartiles, percentiles) by b. Multiplies (divides) measures of spread (range, IQR, standard deviation) by b. Does not change the shape of the distribution. As with data, if we multiply a random variable by a negative constant b, our common measures of spread are multiplied by b. The Practice of Statistics, 5 th Edition 12
Adding a Constant to a Random Variable Consider Pete s Jeep Tours again. We defined C as the amount of money Pete collects on a randomly selected day. Collected c i 300 450 600 750 900 Probability p i 0.15 0.25 0.35 0.20 0.05 The mean of C is $562.50 and the standard deviation is $163.50. It costs Pete $100 per trip to buy permits, gas, and a ferry pass. The random variable V describes the profit Pete makes on a randomly selected day. Profit v i 200 350 500 650 800 Probability p i 0.15 0.25 0.35 0.20 0.05 The mean of V is $462.50 and the standard deviation is $163.50. Compare the shape, center, and spread of the two probability distributions. The Practice of Statistics, 5 th Edition 13
Adding a Constant to a Random Variable Effect on a Random Variable of Adding (or Subtracting) a Constant Adding the same number a (which could be negative) to each value of a random variable: Adds a to measures of center and location (mean, median, quartiles, percentiles). Does not change measures of spread (range, IQR, standard deviation). Does not change the shape of the distribution. The Practice of Statistics, 5 th Edition 14
Linear Transformations Effect on a Linear Transformation on the Mean and Standard Deviation If Y = a + bx is a linear transformation of the random variable X, then The probability distribution of Y has the same shape as the probability distribution of X. µ Y = a + bµ X. σ Y = b σ X (since b could be a negative number). Linear transformations have similar effects on other measures of center or location (median, quartiles, percentiles) and spread (range, IQR). Whether we re dealing with data or random variables, the effects of a linear transformation are the same. Note: These results apply to both discrete and continuous random variables. The Practice of Statistics, 5 th Edition 15
El Dorado Community College El Dorado Community College considers a student to be full-time if he or she is taking between 12 and 18 units. The number of units X that a randomly selected El Dorado Community College full-time student is taking in the fall semester has the following distribution. At El Dorado Community College, the tuition for full-time students is $50 per unit. That is, if T = tuition charge for a randomly selected full-time student, T = 50X. Here is the probability distribution for T and a histogram of the probability distribution: Problem: Find the mean and standard deviation of the probability distribution for T. μ T = 50 14.65 = $732.50 σ T = 50 2.06 = $103.00 The Practice of Statistics, 5 th Edition 16
El Dorado Community College At El Dorado Community College, the tuition for full-time students is $50 per unit. That is, if T = tuition charge for a randomly selected full-time student, T = 50X. Here is the probability distribution for T and a histogram of the probability distribution: In addition to tuition charges, each full-time student at El Dorado Community College is assessed student fees of $100 per semester. If C = overall cost for a randomly selected full-time student, C = 100 + T. Here is the probability distribution for C and a histogram of the probability distribution. Problem: Find the mean and standard deviation of the probability distribution for C. μ C = 100 + 732.50 = $832.50 σ C = $103.00 The Practice of Statistics, 5 th Edition 17
Scaling a test Problem: In a large introductory statistics class, the distribution of raw scores on a test X follows a Normal distribution with a mean of 17.2 and a standard deviation of 3.8. The professor decides to scale the scores by multiplying the raw scores by 4 and adding 10. (a) Define the variable Y to be the scaled score of a randomly selected student from this class. Find the mean and standard deviation of Y. Y = 10 + 4X Y Y 10 4 10 4(17.2) 78.8 X 4 4(3.8) 15.2 X The Practice of Statistics, 5 th Edition 18
Scaling a test Problem: In a large introductory statistics class, the distribution of raw scores on a test X follows a Normal distribution with a mean of 17.2 and a standard deviation of 3.8. The professor decides to scale the scores by multiplying the raw scores by 4 and adding 10. (b) What is the probability that a randomly selected student has a scaled test score of at least 90? Step 1: State the distribution and the values of interest. Because linear transformations do not change the shape, Y follows a N(78.8, 15.2) distribution. We want to find P(Y 90) as shown below. Step 3: Answer the question. There is about a 23% chance that a randomly selected student has a scaled test score of at least 90. The Practice of Statistics, 5 th Edition 19
Transforming and Combining Random Variables Section Summary In this section, we learned how to DESCRIBE the effects of transforming a random variable by adding or subtracting a constant and multiplying or dividing by a constant. The Practice of Statistics, 5 th Edition 20
PAGE 382 36, 38, 42, 46 HOMEWORK The Practice of Statistics, 5 th Edition 21