4.2 (cont.) Standard Deviation of a Discrete Random Variable

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4.2 (cont.) Standard Deviation of a Discrete Random Variable. First center (expected value) Now - spread. 4.2 (cont.) Standard Deviation of a Discrete Random Variable. Measures how “spread out” the random variable is. Data Histogram measure of the center: sample mean x measure of spread:
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4.2 (cont.) Standard Deviation of a Discrete Random VariableFirst center (expected value)Now - spread4.2 (cont.) Standard Deviation of a Discrete Random VariableMeasures how “spread out” the random variable isDataHistogrammeasure of the center: sample mean xmeasure of spread: sample standard deviation sRandom variableProbability Histogrammeasure of the center: population mean mmeasure of spread: population standard deviation sSummarizing data and probabilityExample
  • 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 1/2E(y) = 49(1/2) + 51(1/2) = 50VariationThe deviations of the outcomes from the mean of the probability distribution xi - µ2 (sigma squared) is the variance of the probability distributionVariationVariance of discrete random variable XEconomicScenarioProfit($ Millions)ProbabilityXPGreat100.20x1P(X=x1)5Good0.40x2P(X=x2)OK10.25x3P(X=x3)Lousy-40.15x4P(X=x4)VariationExample2 = (x1-µ)2 · P(X=x1) + (x2-µ)2 · P(X=x2) + (x3-µ)2 · P(X=x3) + (x4-µ)2 · P(X=x4) = (10-3.65)2 · 0.20 + (5-3.65)2 · 0.40 + (1-3.65)2 · 0.25 + (-4-3.65)2 · 0.15 = 19.32753.653.653.653.65P. 207, Handout 4.1, P. 4Standard Deviation: of More Interest then the Variance2 = 19.3275Standard Deviation, or SD, is the standard deviation of the probability distributionProbability Histogram = 4.40µ=3.65Finance and Investment Interpretation
  • X = return on an investment (stock, portfolio, etc.)
  • E(x) = m = expected return on this investment
  • sis a measure of the risk of the investment
  • ExampleA basketball player shoots 3 free throws. P(make) =P(miss)=0.5. Let X = number of free throws made.Expected Value of a Random VariableExample: The probability model for a particular life insurance policy is shown. Find the expected annual payout on a policy.We expect that the insurance company will pay out $200 per policy per year.13© 2010 Pearson Education Standard Deviation of a Random VariableExample: The probability model for a particular life insurance policy is shown. Find the standard deviation of the annual payout.14© 2010 Pearson Education 68-95-99.7 Rule for Random VariablesFor random variables x whose probability histograms are approximately mound-shaped:
  • P(m - s  x  m + s)  .68
  • P(m - 2s  x  m + 2s)  .95
  • P(m -3s  x  m + 3s)  .997
  • (m - 1s, m + 1s) (50-5, 50+5) (45, 55)P(m - s  X  m + s) = P(45  X  55)=.048+.057+.066+.073+.078+.08+.078+.073+ .066+.057+.048=.724Rules for E(X), Var(X) and SD(X):adding a constant a
  • If X is a rv and a is a constant:
  • E(X+a) = E(X)+a
  • Example: a = -1
  • E(X+a)=E(X-1)=E(X)-1
  • Rules for E(X), Var(X) and SD(X): adding constant a (cont.)
  • Var(X+a) = Var(X)
  • SD(X+a) = SD(X)
  • Example: a = -1
  • Var(X+a)=Var(X-1)=Var(X)
  • SD(X+a)=SD(X-1)=SD(X)
  • EconomicScenarioProfit($ Millions)XProbabilityEconomicScenarioProfit($ Millions)X+2ProbabilityPPGreat100.20Great10+20.20x1x1+2P(X=x1)P(X=x1)55+2Good0.40Good0.40x2x2+2P(X=x2)P(X=x2)OK10.25OK1+20.25x3x3+2P(X=x3)P(X=x3)Lousy-40.15Lousy-4+20.15x4x4+2P(X=x4)P(X=x4)E(x + a) = E(x) + a; SD(x + a)=SD(x); let a = 2 = 4.40 = 4.40m=5.65m=3.65New Expected ValueLong (UNC-CH) way: (compute from “scratch”)E(x+2)=12(.20)+7(.40)+3(.25)+(-2)(.15) = 5.65Smart (NCSU) way:a=2; E(x+2) =E(x) + 2 = 3.65 + 2 = 5.65New Variance and SDLong (UNC-CH) way: (compute from “scratch”)Var(X+2)=(12-5.65)2(0.20)+… +(-2+5.65)2(0.15) = 19.3275SD(X+2) = √19.3275 = 4.40Smart (NCSU) way:Var(X+2) = Var(X) = 19.3275SD(X+2) = SD(X) = 4.40Rules for E(X), Var(X) and SD(X): multiplying by constant b
  • E(bX)=b E(X)
  • Var(b X) = b2Var(X)
  • SD(bX)= |b|SD(X)
  • Example: b =-1
  • E(bX)=E(-X)=-E(X)
  • Var(bX)=Var(-1X)=
  • =(-1)2Var(X)=Var(X)
  • SD(bX)=SD(-1X)=
  • =|-1|SD(X)=SD(X)Expected Value and SD of Linear Transformation a + bxLet X=number of repairs a new computer needs each year. Suppose E(X)= 0.20 and SD(X)=0.55The service contract for the computer offers unlimited repairs for $100 per year plus a $25 service charge for each repair.What are the mean and standard deviation of the yearly cost of the service contract?Cost = $100 + $25XE(cost) = E($100+$25X)=$100+$25E(X)=$100+$25*0.20== $100+$5=$105SD(cost)=SD($100+$25X)=SD($25X)=$25*SD(X)=$25*0.55==$13.75Addition 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 independent random variables:
  • Var(X+Y)=Var(X)+Var(Y)
  • SD(X+Y)=
  • SD’s do not add:SD(X+Y)≠ SD(X)+SD(Y)
  • Var(X−Y)=Var(X)+Var(Y)
  • SD(X −Y)=
  • SD’s do not subtract:SD(X−Y)≠ SD(X)−SD(Y)SD(X−Y)≠ SD(X)+SD(Y)Motivation forVar(X-Y)=Var(X)+Var(Y)
  • Let X=amount automatic dispensing machine puts into your 16 oz drink (say at McD’s)
  • A thirsty, broke friend shows up.
  • Let Y=amount you pour into friend’s 8 oz cup
  • Let Z = amount left in your cup; Z = ?
  • Z = X-Y
  • Var(Z) = Var(X-Y) =
  • Var(X)Has 2 components+ Var(Y)Example: rv’s NOT independent
  • X=number of hours a randomly selected student from our class slept between noon yesterday and noon today.
  • Y=number of hours the same randomly selected student from our class was awake between noon yesterday and noon today. Y = 24 – X.
  • What are the expected value and variance of the total hours that a student is asleep and awake between noon yesterday and noon today?
  • Total hours that a student is asleep and awake between noon yesterday and noon 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 add Var(X) and Var(Y) since X and Y are not independent.
  • Pythagorean Theorem of Statistics for Independent X and Ya2+b2=c2Var(X+Y)=Var(X+Y)c2+Var(Y)Var(X)Var(X)ca2aSD(X+Y)SD(X)a + b ≠ cSD(X)+SD(Y) ≠SD(X+Y)bSD(Y)b2Var(Y)Pythagorean Theorem of Statistics for Independent X and Y32 + 42 = 52Var(X)+Var(Y)=Var(X+Y)25=9+16Var(X)Var(X+Y)593SD(X+Y)SD(X)3 + 4 ≠ 5SD(X)+SD(Y) ≠SD(X+Y)4SD(Y)16Var(Y)Example: meal plans
  • Regular plan: X = daily amount spent
  • E(X) = $13.50, SD(X) = $7
  • Expected value and stan. dev. of total spent in 2 consecutive days?
  • E(X1+X2)=E(X1)+E(X2)=$13.50+$13.50=$27
  • SD(X1 + X2) ≠ SD(X1)+SD(X2) = $7+$7=$14Example: meal plans (cont.)
  • Jumbo plan for football players Y=daily amount spent
  • E(Y) = $24.75, SD(Y) = $9.50
  • Amount by which football player’s spending exceeds regular student 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=$2.50For random variables, X+X≠2X
  • Let X be the annual payout on a life insurance policy. From mortality tables E(X)=$200 and SD(X)=$3,867.
  • If the payout amounts are doubled, what are the new expected value and standard deviation?
  • Double payout is 2X. E(2X)=2E(X)=2*$200=$400
  • SD(2X)=2SD(X)=2*$3,867=$7,734
  • Suppose insurance policies are sold to 2 people. The annual payouts are X1 and X2. Assume the 2 people behave independently. What are the expected value and standard deviation of the total payout?
  • E(X1 + X2)=E(X1) + E(X2) = $200 + $200 = $400
  • The risk to the insurance co. when doubling the payout (2X) is not the same as the risk when selling policies to 2 people.
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