Question

In a recent​ year, the mean number of strokes per hole for a famous golfer was approximately 3.9.(a) Find the variance and standard deviation using the fact that the variance of a Poisson distribution is σ2=μ.​(b) How likely is this golfer to play an​ 18-hole round and have more than 72​ strokes?

Answer 1

Variance = 3.9

Standard deviation = 1.975

p(x > 72) = 0.38482

Explanation:

Given that :

Mean number (μ) of strokes per hole is approximately 3.9

For a poisson distribution :

σ² = μ

Hence,

Variance (σ²) = μ = 3.9

The standard deviation σ = √(σ²)

σ = √3.9 = 1.9748417

Poisson distribution relation :

p( x = x) = (λ^x * e^-λ) / x!

For an 18-hole round :

λ = 18 * 3.9 = 70.2

Using the poisson distribution calculator ; where λ = 70.2 ; x = 72

p(x > 72) = 0.38482