Week 4 Wed Sept 22

# Week 4 Wed Sept 22 - WEEK 4 Wednesday Sept 22 Today Cover...

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1 WEEK 4, Wednesday, Sept 22 Today: Cover Sections 3-3, 3-4, 3-5, and start 3-7 . Some Properties of Expectation and of Variance (Sec 3-3). For constants a and b , b a b X aE b aX E ) ( ) ( (3-6) 2 2 2 ) ( ) ( a X V a b aX V (3-10) Example 1. Profit from Products Sold . Suppose that the number X of units of a product sold at a market is a random variable with expectation 57 and standard deviation 4.3. Suppose that the profit Y (in \$) from the sale of X units is given by Y = 10 X – 350. Notice that Y is itself a random variable that is a linear function of the random variable X . The formula (3-6) above allows us to evaluate E ( Y ) knowing the function and E ( X ). Here Y = 10 X – 350 so that E ( Y ) = E (10 X – 350) = 10 E ( X ) – 350 = 10(57) – 350 = \$220. Since Y = 10 X – 350 and the formula (3-10) above shows that 1849 3 . 4 10 ) ( 10 ) 350 10 ( ) ( 2 2 2 X V X V Y V from which SD ( Y ) = = \$43. Summary: Here profit Y is random, expected profit is \$220, and the standard deviation of profit is \$43. Since Y is a linear function of X in this model, we could determine the mean and standard deviation of Y using the formulas (3-6) and (3-10) knowing only the expected value and standard deviation of X .

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2 Example 2. Another Example of Profit from Products Sold . Suppose that the number X of units of a product sold at a market is a random variable. Suppose that the profit Y (in \$) from the sale of X units is given by Y = 100 X 2 – 350. Notice that Y is itself a random variable that is not linear function of the random variable X . The formulas (3-6) and (3- 10) above do not apply. We have to use other ways to determine E ( Y ) and SD ( Y ). Here is a way starting with the probability distribution of X . Suppose the probability distribution of X is as shown below. The tables show how to get the probability distribution of Y and then its mean E ( Y ) and its standard deviation SD ( Y ). x p(x) xp(x) x 2 p(x) y = 100 x 2 - 350 p(y) yp(y) y 2 p(y) 0 0.1 0 0 -350 0.1 -35 12250 1 0.3 0.3 0.3 -250 0.3 -75 18750 2 0.4 0.8 1.6 50 0.4 20 1000 3 0.1 0.3 0.9 550 0.1 55 30250 4 0 0 0 1250 0 0 0 5 0.1 0.5 2.5 2150 0.1 215 462250 Total 1 1.9 5.3 Total 1 180 524500 We see that From the distribution of the random variable Y we see that Remark . The function Y = 100 X 2 - 350 is not a linear function of X so that we used its probability distribution to determine its expectation and standard deviation.
3 Some Results for Sums of Random Variables . ( Did not cover this page in class. Students are responsible for this material and should ask questions if they do not understand it. ) Expectation and variance of a sum . For any constants a 1 , a 2 ,…, a k , b and any random variables X 1 , X 2 ,…, X k , it follows that b X E a X E a X E a b X a X a X a E k k k k ) ( ... ) ( ) ( ) ... ( 2 2 1 1 2 2 1 1 In Sec 3-3. If in addition, the random variables

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## This note was uploaded on 02/01/2011 for the course STAT 315 taught by Professor Dennisgilliland during the Summer '10 term at Michigan State University.

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Week 4 Wed Sept 22 - WEEK 4 Wednesday Sept 22 Today Cover...

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