Question

In a recent​ year, the mean number of strokes per hole for a famous golfer was approximately 3.6. ​(a) Find the variance and standard deviation using the fact that the variance of a Poisson distribution is sigma squared equals mu. ​(b) How likely is this golfer to play an​ 18-hole round and have more than 72​ strokes?

Answer 1

Answer:

a.

Variance=σ²=3.6

Standard deviation=σ=1.897

b.

This golfer is not much likely to play an​ 18-hole round and have more than 72​ strokes because probability is low i.e.16.88%.

Step-by-step explanation:

We are given that mean number of strokes per hole=μ=3.6.

a.

The mean and variance of the Poisson distribution are equal

μ=σ²

Variance=σ²=3.6.

Standard deviation=√variance

σ=√σ²

σ=√3.6

Standard deviation=σ=1.897

b.

P(X>72)=1-P(X≤72)

P(X>72)=1-P(X≤72)

The calculation for this probability are difficult to made by hand and so, excel is used for computation of this probability.

Also we are given the mean number of strokes per hole=μ=3.6 and we have to find the mean number of strokes per 18 hole for computing the probability.

1 hole=3.6

18 hole=3.6*18=64.8

The mean number of strokes per 18 hole=μ=64.8

Using excel function POISSON.DIST(72,64.8,TRUE) for calculating the above probability

P(X>72)=1-0.8312

P(X>72)=0.1688

This golfer is not much likely to play an​ 18-hole round and have more than 72​ strokes because probability is low i.e.16.88%.