mean number of strokes per hole is 3.3

What is the variance?

What is the standard deviation?
 

How likely is the golfer to play 18 hole round and have more than 72 strokes?
asked Feb 27, 2013 in Statistics Answers by anonymous

Your answer

Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:

To avoid this verification in future, please log in or register.

1 Answer

Variance=mean=3.3, SD=sqrt(3.3)=1.82 approx.

Poisson probability P(X)=(e^-µ)µ^X/X!, where µ=mean.

P(72/18)=P(4)=e^-3.3*3.3^4/4!=0.182 is the probability of averaging 72 in 18 holes.

Exclude P(0) in this case because it has no meaning to average zero strokes per hole.

Probability of more than 4 strokes per hole is (1-probability of 1, 2, 3 or 4 strokes per hole).

Probability of more than 4 strokes per hole is 1-(P(1)+P(2)+P(3)+P(4))=

1-e^-3.3(3.3+3.3^2/2+3.3^3/6+3.3^4/24)=

1-3.3e^-3.3(1+3.3/2+3.3^2/6+3.3^3/24)=0.2743 or 27.43%.
answered Mar 9, 2015 by Rod Top Rated User (417,140 points)
Welcome to MathHomeworkAnswers.org, where students, teachers and math enthusiasts can ask and answer any math question. Get help and answers to any math problem including algebra, trigonometry, geometry, calculus, trigonometry, fractions, solving expression, simplifying expressions and more. Get answers to math questions. Help is always 100% free!
77,840 questions
81,454 answers
1,760 comments
60,433 users