Ralph and Claudia are playing a game in which the higher score wins. Ralph\'s sc

Question

Ralph and Claudia are playing a game in which the higher score wins. Ralph's scores in this game are normally distributed with a mean of 100 and a standard deviation of 20. Claudia's are normally distributed with a mean of 110 and a standard deviation of 15.
a) Who is more likely to score above 150?
b) Assuming their scores are independent, what is the approximate probability that Ralph beats Claudia in this format?
c) Suppose they play three games and declare the winner to be the one who gets the highest total for the three games played. What is the probability that Ralph beats Claudia in this format?
d) Why is the mathematics involved so important?
e) This time they play three games but the winner is the one who wins at least two of the three. What is the probability that Ralph beats Claudia in this format?

Explanation / Answer

let X denotes Ralph's score. so X~N(100,202)

Y denotes Claudia's score. so Y~N(110,152)

a) probability that Ralph scores more than 150 is

P[X>150]=P[(X-100)/20>(150-100)/20]=P[Z>2.5] where Z~N(0,1)

=1-P[Z<2.5]=1-0.9937903=0.0062097

probability that Claudia scores more than 150 is

P[Y>150]=P[(Y-110)/15>(150-110)/15]=P[Z>2.667]=1-P[Z<2.667]=1-0.9961734=0.0038266

so P[X>150]>P[Y>150]

hence Ralph is more likely to score more than 150 [answer]

b) the probability that Ralph beats Claudia is

P[X>Y]=P[X-Y>0]

where X-Y~N(100-110,202+152) that is X-Y~N(-10,252) since X and Y are assumed to be independent.

so P[X-Y>0]=P[(X-Y+10)/25>(0+10)/25]=P[Z>0.4]=1-P[Z<0.4]=1-0.6554217=0.3445783 [answer]

c) Suppose they play three games and declare the winner to be the one who gets the highest total for the three games played.

let X1,X2,X3 be the scores of Ralph in the 3 games and Y1,Y2,Y3 be the scores of Claudia in the 3 games

so here X1,X2,X3 are iid N(100,202) and Y1,Y2,Y3 are iid N(110,152)

so X1+X2+X3~N(100+100+100,202+202+202) that is X1+X2+X3~N(300,1200)

Y1+Y2+Y3~N(110*3,152*3) that is Y1+Y2+Y3~N(330,675)

so the probability that Ralph beats Claudia in this format is

P[X1+X2+X3>Y1+Y2+Y3]=P[X1+X2+X3-Y1-Y2-Y3>0]

now X1+X2+X3-Y1-Y2-Y3~N(300-330,1200+675)=N(-30,1875)

so P[X1+X2+X3-Y1-Y2-Y3>0]=P[(X1+X2+X3-Y1-Y2-Y3+30)/sqrt(1875)>(0+30)/sqrt(1875)]

=P[Z>0.6928]=1-P[Z<0.6928]=1-0.7557825=0.2442175 [answer]

e) This time they play three games but the winner is the one who wins at least two of the three.

probability that Ralph beats Claudia in this format=1-probability that Ralph loses to Claudia in this format.

=1-P[Ralph lost all the three matches]-P[Ralph wins only 1 match]

=1-P[Ralph lost all the three matches]-P[Ralph wins 1st match or second match or third match only]

=1-P[X1<Y1,X2<Y2,X3<Y3]-[P[X1>Y1,X2<Y2,X3<Y3]+P[X1<Y1,X2>Y2,X3<Y3]+P[X1<Y1,X2<Y2,X3>Y3]]

now P[X1<Y1]=P[X2<Y2]=P[X3<Y3]=P[X<Y]=P[X-Y<0]=0.6554217 [from b]

P[X1>Y1]=P[X2>Y2]=P[X3>Y3]=P[X>Y]=P[X-Y>0]=1-0.6554217=0.3445783

so required probability is

1-0.65542173-[0.3445783*0.65542172+0.3445783*0.65542172+0.3445783*0.65542172] [since the scores are independent]

=0.274376153 [answer]

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