SOLUTION: You went bowling in two leagues. In the first league the mean bowling score is 167 with a standard deviation of 44. In the second league the mean score is 180 with a standard devia
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-> SOLUTION: You went bowling in two leagues. In the first league the mean bowling score is 167 with a standard deviation of 44. In the second league the mean score is 180 with a standard devia
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Question 1179880: You went bowling in two leagues. In the first league the mean bowling score is 167 with a standard deviation of 44. In the second league the mean score is 180 with a standard deviation of 35. If you bowl a game of 201 in both leagues in which league is this score the best in comparison to the other bowlers? Explain why. Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! normalize the score with z=(x-mean)/sd
for the first z=(201-167)/44=34/44=0.77
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for the second =(201-180)/35=21/35=0.6
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The first z-score is higher which means one is at a higher percentile of the group than the second z=score.
