Quantitative Analysis

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  Question 1For normal distribution with mean = 60 and standard deviation = 6, determine the z = (x - mu)/sigma = (62 - 60)/6 = 0.3333 and z = (67 - 60)/6 = 1.1667P(62 < x < 67) = P(0.3333 < z < 1.1667) = 0.2478 Question 2!he table below ives the deviation of a hypothetical portfolio#s annual total return *ortfolio+s eviation from benchmar& return, 1--2'200('-)1-./.')')..0)0-0)0-.3.)7()7(..-)(7 Class IntervalFrequencyCumulave FrequencyRelave Frequency -9.19<A<-4.55330.25-4.55<B<0.09470.330.09<<4.733100.254.73<!<9.372120.17 Total121.00 $2% 3onstruct a histo ram usin the data) 4dentify the modal interval of the roupe 1--2 '7)151--( 1)6251-- 2)51-- '2)-51--6 -)(751--7 '0)51-- '0)-51--- '-)1-52000 ')1152001 '0)-52002 6)5200( ()05$1% 3alculate the freuency, cumulative freuency, relative freuency, and cum  The modal interval is -.!! to 0.0 Trac#ing Ris# $ %tandard &eviaon o' the deviaons 'rom the (enchmar# return $ ! -9.19<A<-4.55-4.55<B<0.090.09<<4.734.73<!<9.3700.511.522.533.544.5 Histogram Class Interval        F      r      e      q      u      e      n      c      y   probability content of the interval [62 to 67 $ ross of fees% from its benchmar&#s annual returns, for a 12'year period endin in 200() *ear&eviaon 'rom +enchmar# Return ,%orted 1999-9.191992-7.142000-5.111995-2.591998-0.891997-0.552001-0.4919931.6219942.4820033.0420026.8419969.37 %tandard &eviaon s $ !.1Cumulave Relave Frequency 0.250.580.831.00  d data) !rac&in ris& $also called trac&in error% is the standard deviation of the deviation of a portfolio#s lative relative freuency for the portfolio#s deviation from benchmar& return, iven the set of intervals in   .1).
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