Lesson 8 Section 6.2—Linear Transformations and Combining Normal Random Variables

WARM-UP! Given independent random variables with means and standard deviations as shown, SHOW YOUR WORK, and calculate the mean and standard deviation of: MEAN SD a) 0.8Y 0.8 𝜇 𝑌 = 0.8(300) = 240 0.8 𝜎 𝑌 = 0.8(16) = 12.8 b) 2X – 100 2 𝜇 𝑋 −100 = 2(120) - 100 = 140 2 𝜎 𝑋 = 2(12) = 24 (NO “Adder”) c) X + 2Y 𝜇 𝑋 +2 𝜇 𝑌 = 120 + 2(300) = 720 𝜎 𝑋 2 + 2 2 𝜎 𝑌 2 = 12 2 + 4∗16 2 = 34.18 d) 3X – Y 3 𝜇 𝑋 − 𝜇 𝑌 = 3(120)-300 = 60 3 2 𝜎 𝑋 2 + 𝜎 𝑌 2 = 9∗ 12 2 + 16 2 = 39.40 e) 𝑌 1 + 𝑌 2 𝜇 𝑌 + 𝜇 𝑌 = 300 + 300 = 600 𝜎 𝑌 2 + 𝜎 𝑌 2 = 16 2 + 16 2 = 22.63 MEAN SD X 120 12 Y 300 16

More Practice with Linear Transformations Scaling a Test: In a large intro statistics class, the distribution of X = raw test scores was approximately Normally distributed with a mean of 17.2 and a standard deviation of 3.8. The professor decides to scale the scores (aka CURVE the test!) by multiplying the raw scores by four and adding ten. 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—show formulas used:   mean: Since Y = 4X + 10, then 𝜇 𝑌 =4 𝜇 𝑋 +10. Thus, 𝜇 𝑌 = 4(17.2) + 10 = 78.8 SD: Remember, only a “multiplier” will affect the spread, so 𝜎 𝑌= 4 𝜎 𝑋 = 4(3.8) = 15.2

b) What is the probability that a randomly selected student has a scaled score of at least 90? Solution: Since linear transformations do not change the shape of the original distribution, Y is also Normally distributed with N(78.8, 15.2)! So, find 𝑃 𝑌≥90 . Standardize! Calculate the z-score: 𝑧= 90−78.8 15.2 = 0.74 Draw curve and shade the region!

Find the area under the curve using either Table A or the calculator: Using Table A, 𝑃 𝑧≥0.74 =1−0.7704=0.2296 Using technology, 𝑃 𝑧≥0.74 = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 0.74, 𝐸99, 0, 1 =0.2296 Conclude: There is about a 23% chance that a randomly selected student would have a scaled score of at least 90.

Medley Relay In the 4x100 medley relay event, four swimmers swim 100 yards, each using a different stroke. A college team preparing for the conference championship looks at the times their swimmers have posted and creates a model based on the following assumptions: The swimmers’ performances are independent Each swimmer’s time follows a Normal model The means and standard deviations of the times are given below: Swimmer Mean SD Backstroke 50.72 0.24 2) Breaststroke 55.51 0.22 3) Butterfly 49.43 0.25 4) Freestyle 44.91 0.21

Medley Relay 𝜇 𝑇 = 𝜇 1 + 𝜇 2 + 𝜇 3 + 𝜇 4 What are the mean and standard deviation for the relay team’s total time in this event? 𝜇 𝑇 = 𝜇 1 + 𝜇 2 + 𝜇 3 + 𝜇 4 𝜇 𝑇 =50.72+55.51+49.43+44.91 = 200.57 sec 𝜎 𝑇 2 = 𝜎 1 2 + 𝜎 2 2 + 𝜎 3 2 + 𝜎 4 2 𝜎 𝑇 2 = 0.24 2 + 0.22 2 + 0.25 2 + 0.21 2 =0.2126 ∴ 𝜎 𝑇 = 0.2126 =0.46 seconds

Medley Relay The team’s best time so far this season was 3:19.48 (That’s 199.48 seconds.) Do you think the team is likely to swim faster than this at the conference championship meet? Justify your answer! Find 𝑃(𝑇<199.48) *Since each individual swimmer has a Normal distribution, then the distribution of the SUM of all of the swimmers is also Normal! N(200.57, 0.46)

Standardize, draw and shade the curve, find area under the curve: 𝑧= 199.48−200.57 0.46 =−2.37 Using technology, 𝑃 𝑧<−2.37 = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 −𝐸99, −2.37, 0, 1 =0.0089 So, NO, we would not expect them to swim faster since there’s only an 0.89% chance that they would do so. . .but GOOD LUCK TRYING! 

Combining Normal Random Variables If a random variable is Normally distributed, we can use its mean and variance to compute probabilities. What if we combine two Normal random variables? Any linear combination (sum or difference) of independent Normal random variables is also Normally distributed! If X and Y are independent Normal random variables and a and b are any fixed numbers, then aX + bY is also Normally distributed. You must CLEARLY communicate this fact about the distribution’s shape before proceeding any further!

Combining Normal Random Variables EXAMPLE: Tom and George are avid golf players. Their scores vary as they play the course repeatedly according to the following distributions:   Tom’s score, T: N(110, 10) George’s score, G: N(100, 8) If they play independently, what is the probability that Tom will score lower than George (and thus do better in the tournament)? Use the four-step process. Find 𝑃(𝑇<𝐺)!

Four –Step Process State: What is the probability that Tom will have a lower golf score than George? 𝑃(𝑇<𝐺) Plan: Tom’s score is independent of George’s score. Set a variable D to be the difference between their scores, so we won’t have to deal with two variables. . .let D = T – G. Since we are looking for Tom’s score to be lower than George’s score (for Tom to win) then we want D (the difference) to be less than 0, so 𝑃 𝑇−𝐺<0 , or 𝑃 𝐷<0 !

DO: Since T and G are both independent Normal random variables, then D, their difference, follows a Normal distribution with mean 𝜇 𝐷 = 𝜇 𝑇 − 𝜇 𝐺 =110−100=10 and variance 𝜎 𝐷 2 = 𝜎 𝑇 2 + 𝜎 𝐺 2 = (10) 2 + (8) 2 =164 and the standard deviation is the square root of the variance: 𝜎 𝐷 = 164 = 12.8

To find this area, we can use Table A, or use calculator: Thus, “D = T – G” is Normal N(10, 12.8). Draw Normal curve and shade appropriate area for the standardized value, 𝑧= 0−10 12.8 =−0.78125 To find this area, we can use Table A, or use calculator: Using technology, 𝑃 𝑧<−0.78125 = 𝑛𝑜𝑟𝑚𝑎𝑙𝑐𝑑𝑓 −𝐸99, −0.78125, 0, 1 =0.2173 Conclude: There is about a 22% chance that Tom will have a lower score than George and beat George in their golf match.

Linear Transformations of Random Variables—Extra Practice—Part 1 EXAMPLE 1: A large auto dealership keeps track of sales made during each hour of the day. Let X= the number of cars sold during the first hour of business on a randomly selected Friday. The probability distribution of X is: The mean is 𝜇 𝑋 =1.1 ; the standard deviation is 𝜎 𝑋 =0.943 . Suppose the dealership’s manager receives a $500 bonus for each car sold. Let Y = the bonus received. Find 𝜇 𝑌 and 𝜎 𝑌 . 𝜇 𝑌 = (500)(1.1) = $550 𝜎 𝑌 = (500)(0.943) = $471.50 # of Cars Sold (X) 1 2 3 Probability .3 .4 .2 .1

Linear Transformations of Random Variables EXAMPLE: A large auto dealership keeps track of sales made during each hour of the day. Let X= the number of cars sold during the first hour of business on a randomly selected Friday. The probability distribution of X is: The mean is 𝜇 𝑋 =1.1 ; the standard deviation is 𝜎 𝑋 =0.943 . To encourage customers to buy cars, the manager spends $75 to provide coffee and doughnuts. The manager’s net profit, T, on a random Friday is the bonus earned minus $75. Find 𝜇 𝑇 and 𝜎 𝑇 . 𝜇 𝑇 = (500)(1.1) – 75 = $475; 𝜎 𝑇 = (500)(0.943) = $471.50 # of Cars Sold (X) 1 2 3 Probability .3 .4 .2 .1

Combining Random Variables—Part II Example 1: A large auto dealership keeps track of sales and lease agreements made during each hour of the day. Let X = # of cars sold, and Y = # cars leased during the first hour of business on a randomly selected Friday. Based on previous records, the probability distributions of X and Y are as follows: (continued on next slide . . .)

Example continued. . . Cars Sold, xi 1 2 3 Probability, pi 0.3 0.4 0.2 1 2 3 Probability, pi 0.3 0.4 0.2 0.1 Cars Leased, yi 1 2 Probability, pi 0.4 0.5 0.1

Example continued. . . Define T = X + Y. µx = 1.1; σx = 0.943 and µY = 0.7; σY = 0.64 Define T = X + Y. 1) Find and interpret µT. 𝜇 𝑇 = 𝜇 𝑋 + 𝜇 𝑌 = 1.1 + 0.7 = 1.8. On average, this dealership sells or leases 1.8 cars in the first hour of business on Fridays. 2) Compute σT assuming that X and Y are independent. 𝜎 𝑇 = (0.943) 2 + (0.64) 2 = 1.14

Example continued . . . 3) The dealership’s manager receives a $500 bonus for each car sold and a $300 bonus for each car leased. Find the mean and standard deviation of the manager’s total bonus, B. Show work! µB = 500(1.1) + 300(0.7) = $760 σB = 𝟓𝟎𝟎 𝟐 (𝟎.𝟗𝟒𝟑) 𝟐 + (𝟑𝟎𝟎) 𝟐 (𝟎.𝟔𝟒) 𝟐 =$𝟓𝟎𝟗.𝟎𝟗

APPLES! Suppose that a certain variety of apples have weights that are approximately Normally distributed with a mean of 9 oz and a standard deviation of 1.5 oz. If bags of apples are filled by randomly selecting 12 apples, what is the probability that the sum of the weights of the 12 apples is less than 100 oz? P(X<100) = 0.0620

Speed Dating To save time and money, many single people have decided to try speed dating. At a speed-dating event, women sit in a circle, and each man spends about 5 minutes getting to know a woman before moving on to the next one. Suppose that the height M of male speed daters follows a Normal distribution, with a mean of 70” and a standard deviation of 3.5”, and suppose that the height F of female speed daters follows a Normal distribution, with a mean of 65” and a standard deviation of 3”. What is the probability that the man is taller than the woman in a randomly selected speed-dating couple? Use 4-step process! Show all work! P(M>F) = P(D>0), where D = M-F. P(D>0) = 0.8610

Lesson Objectives At the end of the lesson, students can: Perform linear transformations on Random Variables. Determine the mean and standard deviation of transformed Random Variables. Determine the mean and standard deviation of sums and differences of Random Variables.

HOMEWORK Section 6.2, pp. 379-382 DO: #45-51, 59, 63, 65, 66 Work on the WEBASSIGN!