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ARE YOU READY FOR THE QUIZ TODAY 1. 2. 3. Yes 14% 14% No With the little help from a friend 14% 14% 14% 14% 14% Slide 1- 1 1 2 3 4 5 6 7 UPCOMING IN CLASS Quiz 3 Thursday – February 21 (HW4 and HW5) Sunday Homework 6 Exam 1 – March 7th CHAPTER 16 Random Variables EVERYTHING E X x P x Var X x P x 2 2 SD X Var X Slide 1- 4 MORE ABOUT MEANS AND VARIANCES Adding or subtracting a constant from data E(X ± c) = E(X) ± c Var(X ± c) = Var(X) Multiplying each value of a random variable by a constant multiplies the mean by that constant and the variance by the square of the constant: E(aX) = aE(X) Var(aX) = a2Var(X) Slide 1- 5 MORE ABOUT MEANS AND VARIANCES (CONT.) In general, The mean of the sum of two random variables is the sum of the means. The mean of the difference of two random variables is the difference of the means. E(X ± Y) = E(X) ± E(Y) If the random variables are independent, the variance of their sum or difference is always the sum of the variances. Var(X ± Y) = Var(X) + Var(Y) Slide 1- 6 FOR THE FOLLOWING PROBLEMS, Let, Mean of X = 60 SD of X = 10 Mean of Y = 10 SD of Y = 2 Slide 1- 7 FIND THE MEAN AND SD FOR THE RANDOM VARIABLE 4X+2Y 1. 2. 3. 4. 4*60+2*10 and 4*10 4*60+2*10 and 2*2 4*60+2*10 and 4*10+2*2 4*60+2*10 and sqrt(16*100+4*4) 0% 0% 0% 0%Slide 1- 8 1 2 3 4 FIND THE MEAN AND SD FOR THE RANDOM VARIABLE 2X-5Y 1. 2. 3. 4. 2*60-5*10 and sqrt(2*10-5*2) 2*60-5*10 and sqrt(4*10-25*2) 2*60-5*10 and sqrt(4*100-25*4) 2*60-5*10 and sqrt(4*100+25*4) 0% 0% 0% 0%Slide 1- 9 1 2 3 4 FIND THE MEAN AND SD FOR THE RANDOM VARIABLE X1+X2 1. 2. 3. 4. 60+60 and sqrt(100+100) 60+60 and sqrt(10+10) 60+60 and 100+100 60+60 and 10+10 0% 0% 0% 0%Slide 1- 10 1 2 3 4 EGG EXAMPLE A grocery supplier believes that in a dozen eggs, the mean number of broken eggs is 0.5 with a SD of 0.2 eggs. You buy 3 dozen eggs. Slide 1- 11 HOW MANY BROKEN EGGS DO YOU EXPECT IN THE THREE CARTONS? 1. 2. 3. 4. 0.5+0.5+0.5 0.5*0.5*0.5 0.5+0.2 I need a probability model to answer this question 25% 25% 25% 25% Slide 1- 12 1 2 3 4 WHAT’S THE SD OF THE NUMBER OF BROKEN EGG WHEN BUYING THREE CARTONS? 1. 2. 3. 4. 0.2*0.2*0.2 0.2+0.2+0.2 Sqrt(0.2+0.2+0.2) Sqrt(0.22+0.22+0.22) 25% 25% 25% 25% Slide 1- 13 1 2 3 4 WHAT ASSUMPTION DID YOU MAKE ABOUT THE EGGS? 1. 2. 3. 4. The cartons are dependent The number of broken eggs is a continuous random variable. The number of broken eggs is a discrete random variable. The cartons are independent of each other. 25% 25% 25% 25% Slide 1- 14 1 2 3 4 *CORRELATION AND COVARIANCE If X is a random variable with expected value E(X)=µ and Y is a random variable with expected value E(Y)=ν, then the covariance of X and Y is defined as Cov( X , Y ) (( X )(Y )) The covariance measures how X and Y vary together. Slide 1- 15 *CORRELATION AND COVARIANCE (CONT.) Covariance, unlike correlation, doesn’t have to be between -1 and 1. If X and Y have large values, the covariance will be large as well. To fix the “problem” we can divide the covariance by each of the standard deviations to get the correlation: Corr ( X , Y ) Cov( X , Y ) XY Slide 1- 16 INSURANCE POLICIES An insurance company estimates that it should make an annual profit of $130 on each homeowner’s policy, with a standard deviation of $4,000. Slide 1- 17 WHAT IS THE MEAN AND SD IF THE COMPANY WRITE 3 POLICIES? 1. 2. 3. 4. Mean=3*130 SD=sqrt(3*4,000) Mean=130+130+130 SD=sqrt(3*4,0002) Mean=3*130 SD=3*4,000 Mean=130+130+130 SD=3*4,0002 25% 25% 25% 25% Slide 1- 18 1 2 3 4 WHAT IS THE MEAN AND SD IF THE COMPANY WRITE 10,000 POLICIES? 1. 2. 3. 4. Mean=10,000*130 SD=sqrt(10,000*4,0002) Mean=130+130+130 SD=sqrt(10,000*4,0002) Mean=10,000*130 SD=10,000*4,000 Mean=130+130+130 SD=10,000*4,000 25% 25% 25% 25% Slide 1- 19 1 2 3 4 WILL THE COMPANY BE PROFITABLE? 1. 2. 3. 4. No. The variance is larger than the mean. Yes, $0 is 3.25 standard deviations below the mean for 10,000 policies Yes, the expected value is greater than zero. No. Catastrophes are far too unpredictable to expect a profit. 25% 25% 25% 25% Slide 1- 20 1 2 3 4 WHAT ASSUMPTION DID YOU MAKE? 1. 2. 3. 4. The annual profit on a policy is a continuous random variable Losses are dependent Losses are independent of each other The annual profit on a policy is a discrete random variable. 25% 25% 25% 25% Slide 1- 21 1 2 3 4 CEREAL BOWLS Large Bowl E(XL)=2.7 oz SD(XL)=0.2 oz Small Bowl E(XS)=1.7 oz SD(XS)=0.2 oz Slide 1- 22 HOW MUCH MORE CEREAL DO YOU EXPECT TO EAT BY USING THE LARGE BOWL? WHAT IS THE SD? 1. 2. 3. 4. Mean=2.7-1.7 SD=sqrt(0.22-0.22) Mean=2.7-1.7 SD=sqrt(0.22+0.22) Mean=2.7+1.7 SD=sqrt(0.22-0.22) Mean=2.7+1.7 SD=sqrt(0.22+0.22) 25% 25% 25% 25% Slide 1- 23 1 2 3 4 ASSUMING NORMALITY, WHAT IS THE PROB. THAT THE SMALL BOWL CONTAIN MORE CEREAL? 1. 2. 3. 4. -3.57 -3.25 0.0002 0.9998 25% 25% 25% 25% Slide 1- 24 1. 2. 3. 4. WHAT ARE THE MEAN AND SD OF THE TOTAL AMOUNT IN THE TWO BOWLS? 1. 2. 3. 4. Mean=2.7-1.7 SD=sqrt(0.22-0.22) Mean=2.7-1.7 SD=sqrt(0.22+0.22) Mean=2.7+1.7 SD=sqrt(0.22-0.22) Mean=2.7+1.7 SD=sqrt(0.22+0.22) 25% 25% 25% 25% Slide 1- 25 1 2 3 4 ASSUMING NORMALITY, WHAT’S THE PROB. YOU POURED MORE THAN 4.7 OZ OF CEREAL TOTAL? 1. 2. 3. 4. 1.0714 -1.0714 .8577 .1423 25% 25% 25% 25% Slide 1- 26 1. 2. 3. 4. UPCOMING IN CLASS Quiz 3 Thursday – February 21 (HW4 and HW5) Sunday Homework 6 Exam 1 – March 7th