7.2 Part 3 Rules for Means and Variances Students will use the rules for means and variances to solve problems. Linear Transformations Review Multiplying (or dividing) each observation by a constant b (positive, negative, or zero): Multiplies (or divides) measures of center and location (mean, median, quartiles, percentiles) by b Multiplies (or divides) measures of spread (range, IQR, standard deviation) by |b| Does not change the shape of the distribution If you are multiplying or dividing, you are basically stretching or shrinking your distribution Linear Transformations Review Adding or subtracting a constant: Adding the same number a (positive, negative or zero) to each observation: Adds a to measures of center and location (mean, median, quartiles, percentiles) Does not change shape or measures of spread (range, IQR, standard deviation) If adding or subtracting, you are basically shifting your distribution to the left or to the right Rules for Means and Variances 1. Suppose μX = 5 and μY = 10. According to the rules for means, what is μX+Y ? Rule 2 for means states that μX+Y = μX + μY So μX+Y = 5 + 10 = 15 2. Suppose μX = 2. According to the rules for means, what is μ3+4X ? • Rule 1 for means states that μa+bX = a + bμX – So μ3+4X = 3 + 4(2) = 11 3. Suppose σX2 = 2 and σY2 = 3 and X and Y are independent random variables. According to the rules for variances, what is σX2+ Y ? What is σX + Y ? Rule 2 for variances states that 2 X Y 2 So + = 2 X 2 Y 2 X Y 2 X 2 Y 2 + 3 = 5 and + = 5 = 2.23 This rule only applies if X and Y are independent we add the variances whether Keep in mind that we are finding the sum or difference because in both situations we are combining distributions, which gives us more variability 4. Suppose σ2= 4. According to the rules for variances, what is σ23+4x? What is σ3+4x? Rule 1 for variances states that So 2 a bX b 2 2 X σ23+4x = (42)(4) = 64 and σ3+4x = 8. 5. What is the best way to combine standard deviations? • Standard deviations are most easily combined by using the rules for variances. Pete’s Jeep Tours 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. The number of passengers X on a randomly selected day has the following probability distribution: # of passengers X 2 3 4 5 6 Probability 0.15 0.25 0.35 0.20 0.05 Find the mean and standard deviation of X. 1. L1: number of passengers X 2. L2: probability 3. L3: L1 x L2 (The mean is the sum of L3.) 4. L4: (L1-mean)2 5. L5: L4 x L2 (The variance is the sum of L5.) 6. The Standard deviation is the square root of the sum of L5. Pete’s Jeep Tours # of passengers X 2 3 4 5 6 Probability 0.15 0.25 0.35 0.20 0.05 Pete charges $150 per passenger. Let C = the total amount of money that Pete collects on a randomly selected trip. C = 150X The probability distribution of C looks like this: Amount of money collected C $300 $450 $600 $750 $900 Probability 0.15 0.25 0.35 0.20 0.05 Note that the probability doesn’t change Pete’s Jeep Tours # of passengers X 2 3 4 5 6 Probability 0.15 0.25 0.35 0.20 0.05 Amount of money collected C $300 $450 $600 $750 $900 Probability 0.15 0.25 0.35 0.20 0.05 μX = 3.75 σX = 1.090 σ2X = 1.188 μC = $562.50 σC = $163.50 σ2C = $26,718.75 Using the rules, we know that since C = 150X: μC = 150(μX) = 150(3.75) = 562.5 σ2C = 1502(σ2X) = (22,500)(1.188) = 26,730 Or σC = 1.090(150) = 163.5 Pete’s Jeep Tours Amount of money collected C $300 $450 $600 $750 $900 Probability 0.15 0.25 0.35 0.20 0.05 It costs Pete $100 to buy permits, gas, and a ferry pass for each half-day trip. The amount of profit V that Pete makes from the trip is the total amount of money C that he collects from passengers minus $100. That is, V = C – 100. If Pete has only two passengers on the trip (X = 2), then C = 300 and V = 200. V = C – 100 or V = 150X – 100 The probability distribution of V looks like this: Amount of profit V $200 $350 $500 $650 $800 Probability 0.15 0.25 0.35 0.20 0.05 Pete’s Jeep Tours μX = 3.75 σX = 1.090 μC = $562.50 σ2X = 1.188 σC = $163.50 σ2C = $26,718.75 Amount of profit V $200 $350 $500 $650 $800 Probability 0.15 0.25 0.35 0.20 0.05 Using the rules, we know that since V = C – 100: μV = μC – 100 = 562.5 – 100 = 462.5 σ2V and σV stay exactly the same as σ2C and σC because the amount of variability hasn’t increased or decreased μV = $462.50 σV = $163.50 σ2V = $26,718.75 Using the rules, we know that since V = 150X – 100 μV = 150(μX) – 100 = 150(3.75) – 100 = 562.5 – 100 = 462.5 σ2V = 1502(σ2X) = (22,500)(1.188) = 26,730 Or σV = 1.090(150) = 163.5 Erin’s Adventures Pete’s sister Erin, who lives near a tourist area in another part of the country, is impressed by the success of Pete’s business and decides to join the business, running tours on the same days as Pete in her slightly smaller vehicle, under the name “Erin’s Adventures.” After a year of steady bookings, Erin discovers that the number of passengers Y on her half-day tours has the following probability distribution: # of passengers Y 2 3 4 5 Probability 0.3 0.4 0.2 0.1 μY = 3.10 σY = 0.943 σ2Y = .889 Erin’s Adventures # of passengers Y 2 3 4 5 Probability 0.3 0.4 0.2 0.1 μY = 3.10 σY = 0.943 σ2Y = .889 How many total passengers T can Pete and Erin expect to have on their tours on a randomly selected day? Using the rules, we know that since T = X + Y: μT = μX + μY = 3.75 + 3.10 = 6.85 σ2T = σ2X + σ2Y = 1.188 + .889 = 2.077 σT = 2.077 = 1.441 • The probability distribution for T can be found: xi pi yi pi ti = xi + yi pi 2 0.15 2 0.3 4 (0.15)(0.3) = 0.045 2 0.15 3 0.4 5 (0.15)(0.4) = 0.060 2 0.15 4 0.2 6 (0.15)(0.2) = 0.030 2 0.15 5 0.1 7 (0.15)(0.1) = 0.015 3 0.25 2 0.3 5 (0.25)(0.3) = 0.075 3 0.25 3 0.4 6 (0.25)(0.4) = 0.100 3 0.25 4 0.2 7 (0.25)(0.2) = 0.050 3 0.25 5 0.1 8 (0.25)(0.1) = 0.025 4 0.35 2 0.3 6 (0.35)(0.3) = 0.105 4 0.35 3 0.4 7 (0.35)(0.4) = 0.140 4 0.35 4 0.2 8 (0.35)(0.2) = 0.070 4 0.35 5 0.1 9 (0.35)(0.1) = 0.035 5 0.2 2 0.3 7 (0.2)(0.3) = 0.060 5 0.2 3 0.4 8 (0.2)(0.4) = 0.080 5 0.2 4 0.2 9 (0.2)(0.2) = 0.040 5 0.2 5 0.1 10 (0.2)(0.1) = 0.020 6 0.05 2 0.3 8 (0.05)(0.3) = 0.015 6 0.05 3 0.4 9 (0.05)(0.4) = 0.020 6 0.05 4 0.2 10 (0.05)(0.2) = 0.010 6 0.05 5 0.1 11 (0.05)(0.1) = 0.005 Erin’s Adventures The probability distribution for T can be found: Combined # of passengers 4 Probability 0.045 μT = 6.85 5 6 7 8 9 10 11 0.135 0.235 0.265 0.190 0.095 0.030 0.005 σT = 1.441 σ2T = 2.0775 Erin’s Adventures Let D = X – Y, or the difference in the number of passengers that Pete and Erin have on their tours on a randomly selected day. μD = μX – μY = 3.75 – 3.10 = 0.65 σ2D = σ2X + σ2Y = 1.188 + .889 = 2.077 σD = 2.0477 = 1.441 7.41 Time and Motion (pg. 500) (a) A time and motion study measures the time required for an assembly-line worker to perform a repetitive task. The data show that the time required to bring a part from a bin to its position on an automobile chassis varies from car to car with mean 11 seconds and standard deviation 2 seconds. The time required to attach the part to the chassis varies with mean 20 seconds and standard deviation 4 seconds. What is the mean time required for the entire operation of positioning and attaching the part? 7.41 Time and Motion (pg. 500) (b) If the variation in the worker’s performance is reduced by better training, the standard deviations will decrease. Will this decrease change the mean you found in (a) if the mean times for the two steps remain as before? (c) The study finds that the times required for the two steps are independent. A part that takes a long time to position, for example, does not take more or less time to attach than other parts. Find the standard deviation of the time required for the two-step assembly operation. 7.42 Electronic circuit (pg. 500) The design of an electronic circuit calls for a 100-ohm resistor and a 250-ohm resistor connected in series so that their resistances add. The components used are not perfectly uniform, so that the actual resistances vary independently according to Normal distributions. The resistance of 100-ohm resistors has mean 100 ohms and standard deviation 2.5 ohms, while that of 250-ohm resistors has mean 250 ohms and standard deviation 2.8 ohms. (a) What is the distribution of the total resistance of the two components in series? (b) What is the probability that the total resistance lies between 345 and 355 ohms? HOMEWORK 7.38-40, 42, 45, 46 (show work for b, you can use your calculator to calculate expected value for a), 47, 49-52

© Copyright 2019 ExploreDoc