Question

1. last year, at rolling greens golf course, golfers had a mean score of 53 and a

standard deviation of 2.3 on the first nine holes. on the back nine, the golfers had a mean
score of 59 and a standard deviation of 1.8. during her first round of golf in the spring,
kirstyn scored 51 on the front nine and 57 on the back nine. on which half of the course
did kirstyn golf better, relative to the others golfers? (5 marks)

Explanation:

Step1: Calculate the z - score for the front nine

The z - score formula is \(z=\frac{x-\mu}{\sigma}\). For the front nine, \(x = 51\), \(\mu=53\), \(\sigma = 2.3\).

$$z_{front}=\frac{51 - 53}{2.3}=\frac{- 2}{2.3}\approx - 0.87$$

Step2: Calculate the z - score for the back nine

For the back nine, \(x = 57\), \(\mu = 59\), \(\sigma=1.8\)

$$z_{back}=\frac{57 - 59}{1.8}=\frac{-2}{1.8}\approx - 1.11$$

Answer:

Since \(-0.87>-1.11\) (because \(\vert-0.87\vert = 0.87\) and \(\vert-1.11\vert=1.11\) and for negative z - scores, the one with a smaller magnitude is better in terms of relative performance), Kirstyn did better on the front nine.