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review the case During the in PTER S Joint Probability Distributions and Random

ID: 3131307 • Letter: R

Question

review the case During the in PTER S Joint Probability Distributions and Random Samples A certain automobile manufacturer equips a particular model with either a engine or a four-cylinder engine. Let X and X, be fuel efficiencies fonder I with either a six EXAMPLE 5.31 and randomly selected six- cylinder and four cylinder cars, respectively. With gns h"26..-1.2, and ,-1.5, for i ly, with ,- with 74 1 22 q.x, = V3.69 = 1.92 If we relabel so that Xi refers to the four-cylinder car, then EA, X) = 4, but the variance of the difference is still 3.69. The Case of Normal Random Variables When the X's form a random sample from a normal distribution, X and T, are both normally distributed. Here is a more general result concerning linear combinations ferent means andfor variances), then any linear combination o has a normal distribution. In narticular rv's (with possibly dif y dif. bination of the X;'s also

Explanation / Answer

X1 and X2 denote the fuel efficiencies of the six cylinders cars

X3,X4,X5 denote the fuel efficiencies of the four cylinders cars

from example 5.31 it is given that

X1,X2,X3,X4,X5 all follows normal distributions independently

with E[X1]=E[X2]=26   and E[X3]=E[X4]=E[X5]=22

and V[X1]=V[X2]=1.52   and V[X3]=V[X4]=V[X5]=1.22

question 60.

hence Y=(X1+X2)/2-(X3+X4+X5)/3

so Y being a linear combinations of X1,X2,X3,X$,X5 which are normal distributions

hence T also follows a normal distribution with

E[Y]=E[(X1+X2)/2-(X3+X4+X5)/3]=E[X1]/2+E[X2]/2-E[X3]/3-E[X4]/3-E[X5]/3=26/2+26/2-22/3-22/3-22/3=4

and V[Y]=V[(X1+X2)/2-(X3+X4+X5)/3]=V[X1]/4+V[X2]/4+V[X3]/9+V[X4]/9+V[X5]/9

             =1.52/4+1.52/4+1.22/9+1.22/9+1.22/9=1.605

hence Y~N(4,1.605)

so P[0<=Y]=P[(0-4)/sqrt(1.605)<=(Y-4)/sqrt(1.605)]=P[Z>=-3.157] where Z=(Y-4)/sqrt(1.605)~N(0,1)

                  =1-P[Z<=-3.157]=1-0.0007970=0.999203 [answer]

P[Y>-2]=P[(Y-4)/sqrt(1.605)>(-2-4)/sqrt(1.605)]=P[Z>-4.736]=1-P[Z<-4.736]=1-0.0000011=0.9999989 [answer]