Question

jack and jill are both enthusiastic players of a certain computer game. over the past year, jacks mean score when playing the game is 12,400 with a standard deviation of 1500. during the same period, jills mean score is 14,200, with a standard deviation of 2000. they devise a fair contest: each one will play the game once, and they will compare z - scores. jack gets a score of 14,000, and jill gets a score of 16,000. who won the contest, and what were each of their z - scores?

jacks z = 1.07; jills z = 1.11; jill wins the contest

jacks z = 0.81; jills z = 0.99; jill wins the contest

jacks z = 0.94; jills z = 1.11; jill wins the contest

jacks z = 1.07; jills z = 0.90; jack wins the contest

jacks z = 0.94; jills z = 0.90; jack wins the contest

Explanation:

Step1: Calculate Jack's z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(x\) is the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
For Jack: \(\mu = 12400\), \(\sigma=1500\), \(x = 14000\)
\(z_{Jack}=\frac{14000 - 12400}{1500}=\frac{1600}{1500}\approx1.07\)

Step2: Calculate Jill's z - score

For Jill: \(\mu = 14200\), \(\sigma = 2000\), \(x = 16000\)
\(z_{Jill}=\frac{16000-14200}{2000}=\frac{1800}{2000}=0.90\)

Step3: Compare the z - scores

Since \(z_{Jack}\approx1.07\) and \(z_{Jill}=0.90\), and \(1.07>0.90\)

Answer:

Jack's \(z = 1.07\); Jill's \(z = 0.90\); Jack wins the contest