Variance and Standard Deviation - Penn Math

1 Variance and Standard DeviationChristopher CrokeUniversity of PennsylvaniaMath 115 UPenn, Fall 2011 Christopher CrokeCalculus 115 VarianceThe first first important number describing a probabilitydistribution is the mean or expected valueE(X).The next one is thevarianceVar(X) = 2(X). The square root ofthe Variance is called theStandard (xi) is the probability distribution function for a randomvariable with range{x1,x2,x3,..}and mean =E(X) then:Var(X) = 2= (x1 )2f(x1) + (x2 )2f(x2) + (x3 )2f(x3) +..It is a description of how the distribution spreads .NoteVar(X) =E((X )2).The Standard Deviation has the same units asX. ( ifXismeasured in feet then so is .)Christopher CrokeCalculus 115 VarianceThe first first important number describing a probabilitydistribution is the mean or expected valueE(X).The next one is thevarianceVar(X) = 2(X). The square root ofthe Variance is called theStandard (xi) is the probability distribution function for a randomvariable with range{x1,x2,x3.}

2 }and mean =E(X) then:Var(X) = 2= (x1 )2f(x1) + (x2 )2f(x2) + (x3 )2f(x3) +..It is a description of how the distribution spreads .NoteVar(X) =E((X )2).The Standard Deviation has the same units asX. ( ifXismeasured in feet then so is .)Christopher CrokeCalculus 115 VarianceThe first first important number describing a probabilitydistribution is the mean or expected valueE(X).The next one is thevarianceVar(X) = 2(X). The square root ofthe Variance is called theStandard (xi) is the probability distribution function for a randomvariable with range{x1,x2,x3,..}and mean =E(X) then:Var(X) = 2= (x1 )2f(x1) + (x2 )2f(x2) + (x3 )2f(x3) +..It is a description of how the distribution spreads .NoteVar(X) =E((X )2).The Standard Deviation has the same units asX. ( ifXismeasured in feet then so is .)Christopher CrokeCalculus 115 VarianceThe first first important number describing a probabilitydistribution is the mean or expected valueE(X).

3 The next one is thevarianceVar(X) = 2(X). The square root ofthe Variance is called theStandard (xi) is the probability distribution function for a randomvariable with range{x1,x2,x3,..}and mean =E(X) then:Var(X) = 2= (x1 )2f(x1) + (x2 )2f(x2) + (x3 )2f(x3) +..It is a description of how the distribution spreads .NoteVar(X) =E((X )2).The Standard Deviation has the same units asX. ( ifXismeasured in feet then so is .)Christopher CrokeCalculus 115 VarianceThe first first important number describing a probabilitydistribution is the mean or expected valueE(X).The next one is thevarianceVar(X) = 2(X). The square root ofthe Variance is called theStandard (xi) is the probability distribution function for a randomvariable with range{x1,x2,x3,..}and mean =E(X) then:Var(X) = 2= (x1 )2f(x1) + (x2 )2f(x2) + (x3 )2f(x3) +..It is a description of how the distribution spreads.

4 NoteVar(X) =E((X )2).The Standard Deviation has the same units asX. ( ifXismeasured in feet then so is .)Christopher CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?

5 Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?

6 Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?

7 Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?

8 Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 Problem:Remember the game where players pick balls from anurn with 4 white and 2 red balls. The first player is paid $2 if hewins but the second player gets $3 if she wins. No one gets payedif 4 white balls are have seen that the payout and probabilities for the first playerare:Payout Probability28150115 3615 The expected value was = is the Variance ?Alternative formula for Variance : 2=E(X2) ((X )2) =E(X2 2 X+ 2)=E(X2) +E( 2 X) +E( 2)=E(X2) 2 E(X) + 2=E(X2) 2 2+ 2=E(X2) CrokeCalculus 115 This often makes it easier to compute since we can compute andE(X2) at the same :ComputeE(X) andVar(X) whereXis a randomvariable with probability given by the chart below:XPr(X=x) will see later that forXa binomial random variable 2= :What is the Variance of the number of hits for ourbatter that bats.

9 300 and comes to the plate 4 times?Christopher CrokeCalculus 115 This often makes it easier to compute since we can compute andE(X2) at the same :ComputeE(X) andVar(X) whereXis a randomvariable with probability given by the chart below:XPr(X=x) will see later that forXa binomial random variable 2= :What is the Variance of the number of hits for ourbatter that bats .300 and comes to the plate 4 times?Christopher CrokeCalculus 115 This often makes it easier to compute since we can compute andE(X2) at the same :ComputeE(X) andVar(X) whereXis a randomvariable with probability given by the chart below:XPr(X=x) will see later that forXa binomial random variable 2= :What is the Variance of the number of hits for ourbatter that bats .300 and comes to the plate 4 times?Christopher CrokeCalculus 115 This often makes it easier to compute since we can compute andE(X2) at the same :ComputeE(X) andVar(X) whereXis a randomvariable with probability given by the chart below:XPr(X=x) will see later that forXa binomial random variable 2= :What is the Variance of the number of hits for ourbatter that bats.

10 300 and comes to the plate 4 times?Christopher CrokeCalculus 115 For continuous random variableXwith probability density functionf(x) defined on [A,B] we saw:E(X) = BAxf(x)dx.(note that this does not always exist ifB= .)The Variance will be: 2(X) =Var(X) =E((X E(X))2) = BA(x E(X))2f(x) :(uniform probability on an interval) LetXbe the randomvariable you get when you randomly choose a point in [0,B].a) find the probability density ) find the cumulative distribution functionF(x).c) findE(X) andVar(X) = CrokeCalculus 115 For continuous random variableXwith probability density functionf(x) defined on [A,B] we saw:E(X) = BAxf(x)dx.(note that this does not always exist ifB= .)The Variance will be: 2(X) =Var(X) =E((X E(X))2) = BA(x E(X))2f(x) :(uniform probability on an interval) LetXbe the randomvariable you get when you randomly choose a point in [0,B].a) find the probability density ) find the cumulative distribution functionF(x).