Question

The huge market that exists for golf balls can be partially attributed to the fact that golfers lose them at an average rate of 3 per 18-hole round. Assume that the number of golf balls lost in an 18-hole round is distributed as a Poisson random variable. What is the probability that: No balls will be lost in an 18-hole round? 5 or fewer will be lost in an 18-hole round? 2 or fewer will be lost in a 9-hole round? 3 or fewer will be lost in a 36-hole round?

Answer 1

(a)

Poisson Distribution with = 3 is given by:

                      for x = 0, 1,...

So,

Answer is:

0.0498

(b)

P(X5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) +P(X=4) +P(X=5)

So,

P(X5) = 0.9161

So,

Answer is:

0.9161

(c)

Poisson Distribution with = 1.5 is given by:

                      for x = 0, 1,...

P(X2) = P(X =0) + P(X = 1) + P(X = 2)

So,

P(X2)= 0.8089

So,

Answer is:

0.8089

(d)

Poisson Distribution with = 6 is given by:

                      for x = 0, 1,...

P(X3) = P(X =0) + P(X = 1) + P(X = 2) + P(X=3)

So,

P(X3) = 0.1512

So,

Answer is:

0.1512