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# 4.2 cont. Standard Deviation of a Discrete Random Variable .

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4.2 (cont.) Standard Deviation of a Discrete Random Variable. Measures how
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﻿4.2 (cont.) Standard Deviation of a Discrete Random Variable First focus (expected esteem) Now - spread

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4.2 (cont.) Standard Deviation of a Discrete Random Variable Measures how "spread out" the irregular variable is

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Data Histogram measure of the middle: specimen mean x measure of spread: sample standard deviation s Random variable Probability Histogram measure of the inside: populace mean m measure of spread: populace standard deviation s Summarizing information and likelihood

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Example x 0 100 p(x) 1/2 1/2 E(x) = 0(1/2) + 100(1/2) = 50 y 49 51 p(y) 1/2 1/2 E(y) = 49(1/2) + 51(1/2) = 50

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Variation The deviations of the results from the mean of the likelihood dissemination x i - µ  2 (sigma squared) is the fluctuation of the likelihood circulation

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Variation Variance of discrete arbitrary variable X

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Economic Scenario Profit (\$ Millions) Probability X P Great 10 0.20 x 1 P(X=x 1 ) 5 Good 0.40 x 2 P(X=x 2 ) OK 1 0.25 x 3 P(X=x 3 ) Lousy - 4 0.15 x 4 P(X=x 4 ) Variation Example  2 = (x 1 - µ) 2 · P(X=x 1 ) + (x 2 - µ) 2 · P(X=x 2 ) + (x 3 - µ) 2 · P(X=x 3 ) + (x 4 - µ) 2 · P(X=x 4 ) = (10-3.65) 2 · 0.20 + (5-3.65) 2 · 0.40 + (1-3.65) 2 · 0.25 + (- 4-3.65) 2 · 0.15 = 19.3275 3.65 P. 207, Handout 4.1, P. 4

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Standard Deviation: of More Interest then the Variance

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 2 = 19.3275 Standard Deviation , or SD, is the standard deviation of the likelihood dispersion

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Probability Histogram  = 4.40 µ=3.65

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Finance and Investment Interpretation X = return on a venture (stock, portfolio, and so on.) E(x) = m = expected profit for this speculation s is a measure of the danger of the venture

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Example A ball player shoots 3 free tosses. P(make) =P(miss)=0.5. Let X = number of free tosses made.

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Expected Value of a Random Variable Example : The likelihood display for a specific disaster protection arrangement is appeared. Locate the normal yearly payout on an arrangement. We expect that the insurance agency will pay out \$200 per arrangement every year. 13 © 2010 Pearson Education

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Standard Deviation of a Random Variable Example : The likelihood display for a specific life coverage approach is appeared. Locate the standard deviation of the yearly payout. 14 © 2010 Pearson Education

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68-95-99.7 Rule for Random Variables For irregular factors x whose likelihood histograms are roughly hill molded: P (m - s  x  m + s)  .68 P (m - 2s  x  m + 2s)  .95 P( m - 3s  x  m + 3s)  .997

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( m - 1s, m + 1s ) (50-5, 50+5) (45, 55) P (m - s  X  m + s) = P(45  X  55) =.048+.057+.066+.073+.078+.08+.078+.073+ .066+.057+.048=.724

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Rules for E(X), Var(X) and SD(X) : including a steady an If X is a rv and a will be a consistent: E(X+ a ) = E(X)+ an Example: a = - 1 E(X+ a )=E(X-1)=E(X)- 1

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Rules for E(X), Var(X) and SD(X): including consistent a (cont.) Var(X+ a ) = Var(X) SD(X+ a ) = SD(X) Example: a = - 1 Var(X+ a )=Var(X-1)=Var(X) SD(X+ a )=SD(X-1)=SD(X)

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Economic Scenario Profit (\$ Millions) X Probability Economic Scenario Profit (\$ Millions) X+2 Probability P Great 10 0.20 Great 10+2 0.20 x 1 x 1 + 2 P(X=x 1 ) P(X=x 1 ) 5 5+2 Good 0.40 Good 0.40 x 2 x 2 +2 P(X=x 2 ) P(X=x 2 ) OK 1 0.25 OK 1+2 0.25 x 3 x 3 +2 P(X=x 3 ) P(X=x 3 ) Lousy - 4 0.15 Lousy - 4+2 0.15 x 4 x 4 +2 P(X=x 4 ) P(X=x 4 ) E(x + a ) = E(x) + a ; SD(x + a )=SD(x); let a = 2  = 4.40  = 4.40 m= 5.65 m= 3.65

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New Expected Value Long (UNC-CH) route: (register "starting with no outside help") E(x+2)=12(.20)+7(.40)+3(.25)+(- 2)(.15) = 5.65 Smart (NCSU) way: a =2; E(x+2) =E(x) + 2 = 3.65 + 2 = 5.65

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New Variance and SD Long (UNC-CH) path: (process "sans preparation") Var(X+2)=(12-5.65) 2 (0.20)+… +(- 2+5.65) 2 (0.15) = 19.3275 SD(X+2) = √19.3275 = 4.40 Smart (NCSU) way: Var(X+2) = Var(X) = 19.3275 SD(X+2) = SD(X) = 4.40

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Rules for E(X), Var(X) and SD(X): increasing by consistent b E( b X)= b E(X) Var(b X) = b 2 Var(X) SD(b X) = | b |SD(X) Example: b =-1 E( b X)=E(- X)=-E(X) Var(bX )=Var(- 1X)= =(- 1) 2 Var(X)=Var(X) SD(b X)=SD(- 1X)= =|-1|SD(X)=SD(X)

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Expected Value and SD of Linear Transformation a + b x Let X=number of repairs another PC needs every year. Assume E(X)= 0.20 and SD(X)=0.55 The administration contract for the PC offers boundless repairs for \$100 every year in addition to a \$25 benefit charge for every repair. What are the mean and standard deviation of the yearly cost of the administration contract? Taken a toll = \$100 + \$25X E(cost) = E(\$100+\$25X)=\$100+\$25E(X)=\$100+\$25*0.20= = \$100+\$5= \$105 SD(cost)=SD(\$100+\$25X)=SD(\$25X)=\$25*SD(X)=\$25*0.55= = \$13.75

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Addition and Subtraction Rules for Random Variables E(X+Y) = E(X) + E(Y); E(X-Y) = E(X) - E(Y) When X and Y are autonomous arbitrary factors: Var (X+Y)= Var (X)+ Var (Y) SD(X+Y)= SD\'s don\'t include: SD(X+Y)≠ SD(X)+SD(Y) Var (X−Y)= Var (X) + Var (Y) SD(X −Y )= SD\'s don\'t subtract: SD(X−Y)≠ SD(X)−SD(Y) SD(X−Y)≠ SD(X)+SD(Y)

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Motivation for Var(X-Y)=Var(X) + Var(Y) Let X=amount programmed apportioning machine puts into your 16 oz drink (say at McD\'s) A parched, split companion appears. Let Y=amount you immerse companion\'s 8 oz container Let Z = sum left in your glass; Z = ? Z = X-Y Var(Z) = Var(X-Y) = Var(X) Has 2 parts + Var(Y)

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Example: rv\'s NOT autonomous X=number of hours a haphazardly chose understudy from our class dozed between twelve yesterday and twelve today. Y=number of hours the same arbitrarily chose understudy from our class was wakeful between twelve yesterday and twelve today. Y = 24 – X. What are the normal esteem and difference of the aggregate hours that an understudy is snoozing and alert between twelve yesterday and twelve today? Add up to hours that an understudy is snoozing and conscious between twelve yesterday and twelve today = X+Y E(X+Y) = E(X+24-X) = E(24) = 24 Var(X+Y) = Var(X+24-X) = Var(24) = 0. We don\'t include Var(X) and Var(Y) since X and Y are not autonomous .

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Pythagorean Theorem of Statistics for Independent X and Y a 2 +b 2 =c 2 Var(X+Y) =Var(X+Y ) c 2 +Var(Y ) Var(X) Var(X) c a 2 a SD(X+Y) SD(X) a + b ≠ c SD(X)+SD(Y) ≠SD(X+Y) b SD(Y) b 2 Var(Y)

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Pythagorean Theorem of Statistics for Independent X and Y 3 2 + 4 2 = 5 2 Var(X)+Var(Y)=Var(X+Y) 25=9+16 Var(X) Var(X+Y) 5 9 3 SD(X+Y) SD(X) 3 + 4 ≠ 5 SD(X)+SD(Y) ≠SD(X+Y) 4 SD(Y) 16 Var(Y)

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Example: supper arranges Regular arrangement: X = day by day sum spent E(X) = \$13.50, SD(X) = \$7 Expected esteem and stan. dev. of aggregate spent in 2 back to back days? E(X 1 +X 2 )=E(X 1 )+E(X 2 )=\$13.50+\$13.50=\$27 SD(X 1 + X 2 ) ≠ SD(X 1 )+SD(X 2 ) = \$7+\$7=\$14

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Example: feast arranges (cont.) Jumbo arrangement for football players Y=daily sum spent E(Y) = \$24.75, SD(Y) = \$9.50 Amount by which football player\'s spending surpasses customary understudy spending is Y-X E(Y-X)=E(Y)–E(X)=\$24.75-\$13.50=\$11.25 SD(Y ̶ X) ≠ SD(Y) ̶ SD(X) = \$9.50 ̶ \$7=\$2.50

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For irregular factors, X+X≠2X Let X be the yearly payout on an extra security strategy. From mortality tables E(X)=\$200 and SD(X)=\$3,867. In the event that the payout sums are multiplied, what are the new expected esteem and standard deviation? Twofold payout is 2X . E(2X)=2E(X)=2*\$200= \$400 SD(2X)=2SD(X)=2*\$3,867= \$7,734 Suppose protection approaches are sold to 2 individuals. The yearly payouts are X 1 and X 2 . Expect the 2 individuals act autonomously. What are the normal esteem and standard deviation of the aggregate payout? E(X 1 + X 2 )=E(X 1 ) + E(X 2 ) = \$200 + \$200 = \$400 The hazard to the protection co. when multiplying the payout (2X) is not the same as the hazard when offering arrangements to 2 individuals.

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