In a recent year, the mean number of strokes per hole for a famous golfer was ap

Question

In a recent year, the mean number of strokes per hole for a famous golfer was approximately 3.5
(a) Find the variance and standard deviation using the fact that the variance of a Poisson distribution is
sigma squared equals mu2=.
(b) How likely is this golfer to play an 18-hole round and have more than 72 strokes?
(a) The variance is 3.5
(Round to the nearest tenth as needed.)
The standard deviation is 1.9
strokes. (Round to the nearest tenth as needed.)
(b) The probability of more than 72 strokes in an 18-hole round is
(Round to the nearest thousandth as needed.)

Explanation / Answer

Solution:-

a) If x has a poisson distribution with mean = , Variance (x) =   (Mean = variance) And always

standad deviation = sd = V(x) (for every distribution)  

=> Variance V(x) = = 3.5 sd = V(x) = 3.5 = 1.9

b)

(a) Y has a Poisson distribution with mean =

Mean number of strokes for 18 holes = n* = 18(3.5) = 63

The probability of more than 72 strokes in an 18-hole round is:  0.117

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