Critical thinking? We're in trouble. It's hard for me to tell what grade this is.
Stem and leaf is fine but let's just write out the data.
Rich: 35 37 41 42 43 44 45 48 50 55 n=10
Will: 42 43 44 44 46 47 47 48 49 50 n=10
Use measures of center and measures of variation ...
The three measures of center typically used are the mean, the median and the mode.
Rich's sum = 440 so Rich's mean = 440/10 = 44
Rich's middle values are 43 and 44 so Rich's median = 43.5
All Rich's scores are unique so he doesn't really have a mode.
Will's sum = 460, Will's mean = 460/10 = 46
Will's middle numbers are 46 and 47 so median 46.5
Will's data has two modes tied, 44 and 47.
OK, so far these say Rich is a better golfer; he has lower mean and median scores than Will. (Low score is better in golf.)
Let's look at measures of variation. I'm guessing that they don't want to you calculate standard deviation. So we'll do range and mean absolute deviation.
The range is max-min.
Rich's range is 55-35=20
Will's range is 50-42=8
Will is more consistent.
Rich's mean is 44, his scores are here; let's calculate absolute deviation:
Score 35 37 41 42 43 44 45 48 50 55
AbsDev 9 7 3 2 1 0 1 4 6 11
Rich's deviations add to 44 so his MAD is 4.4
Will's mean is 46
Score: 42 43 44 44 46 47 47 48 49 50
AbsDev 4 3 2 2 0 1 1 2 3 4
The sum is 22, Will's MAD is 2.2
Based on the measures of variation, Will is a more consistent golfer. He's on average worse than Rich, but he always manages to stay pretty close to his average while Rich is all over the place.
Still I'd go with Rich for the tournament.
