Golfer Dave practices put in from about 8 feet away from the hole. He makes abou

Question

Golfer Dave practices put in from about 8 feet away from the hole. He makes about 40% of these putts Dave has no memory from shot to shot, so all precious putts have no effect on his current attempt (a) If Dave attempts 18 putts, what is the probability that he makes exact 10 of them? (b) If Dave attempts 8 putts, what is the probability that he makes at least 6 of them? (c) If Dave at tempts 4 putts what is the mean and standard deviation of his number of made putts? (d) Amy is also a golfer, but when she practices putts, her accuracy improves as she practices. On her first putt, she is successful 30% of the time Her probability of increase goes up by 10% each putt, so her success rate on the second putt is 40%, the third putt in 50% and no forth Over 4 putts, does Amy haw a higher expected number of putts made than Dave Why?

Explanation / Answer

here we use binomial distribution with parameter

n=number of attempt

p=probability of putt

and for binomial distribution Binomial distribution ,P(X=r)=nCrpr(1-p)n-r  

(1) here n=18, p=0.4, r=10

P(X=10)=0.0771 ( using ms-excel command =BINOMDIST(10,18,0.4,0) )

(2) here n=18, p=0.4, r=10

P(X<=5)=0.9502 ( using ms-excel command =BINOMDIST(5,18,0.4,1) )

P(X=at least 6)=P(X>=6)=1-P(X<6)=1-P(X<=5)=1-0.9502=0.0498

(3) here p=0.4 and n=4

mean=np=4*0.4=1.6 and

variance=np(1-p)=4*0.4*(1-0.4)=0.96

standard deviaation=sqrt(variance)=sqrt(0.96)=0.9798

(4) Expected number of put for Amy=1*0.4+1*0.5+1*0.6+1*0.7=2.2

here expected number=sum(ni*pi)

yes Amy have higher expected number=2.2 than Dave=1.6

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