1. Two friends are going bowling. They have different skill levels, with bowling

Question

1. Two friends are going bowling. They have different skill levels, with bowling . Player 1's scores are normally distributed with a mean of 110 and a variance . Player 2's scores are normally distributed with a mean of 140 and a variance Let Xi be the random variable that denotes the bowling score for player i, and scores that are random variables with the following distributions: of 75. of 60 assume the players bowling scores are independent from each other. Answer the (a) What is the distribution of the average of the two bowling scores? (Name (b) What is the probability that the average score of the group will be between questions below: the distribution and specify any parameters of the distribution.) 120 and 140 points? (c) What is the probability that both players bowl a score less than 135?

Explanation / Answer

here let average score is X =(X1+X2)/2 where X1 and X2 are random scores from player 1 and player 2.

as for distribution of scores of player 1 and player 2 both are normal therefore distriution of X is also normal with

mean of X =E(X)=(1/2)*(E(X1)+(E(X2)) =(1/2)*(110+140)=125

and variance of X =(1/2)2 *(Var(X1)+Var(X2)) =(1/4)*(75+60)=33.75

hence std deviation of X =(33.75)1/2 =5.81

from above distribution of X is normal wih N~ (125, 5.812)

b)

probability that average score is between 120 and 140 =P(120<X<140)=P((120-125)/5.81<Z<(140-125)/5.81)

=P(-0.8607<Z<2.5820)=0.9951-0.1947 =0.8004

c) probability that both player score less than 135 =P(1st player has score less then 135)*P(2nd player has score less then 135) =(P(X1<135))*(P(X2<135))=(P(Z<(135-110)/(75)1/2)*(P(Z<(135-140)/(60)1/2)

=(P(Z<2.8868))*(P(Z<-0.6455))= 0.9981 *0.2593 =0.2588

please revert for any clariification required,

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