When Mike goes golfing, his score can be shown by this distrbution: X~N(86.2,5.3). What is the percent chance will his score be in the 80's? What is the probability that his score will be lower than the par (72)?

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When Mike goes golfing his score is can be represented as a normal distribution with mean 86.2 and standard deviation 5.3.

His score will lie in the 80s if it is between 80 and 90. The z-score for 80 is (80 - 86.2)/5.3 = -1.16 The corresponding percentage of area for -1.16 is 0.12302. The z-score for 90 is 0.716 and the corresponding value from the normal distribution table is .76115.

This gives the percent chance that the score lies in the 80s as 0.76115 - 0.12302 = 63.81%

The z-score for 72 is -2.67. The area under the curve for this value is 0.00379. There is a 0.379% chance that his score is below 72.

The percent chance that his score is in the 80s is 63.81% and the chance of the score being below 72 is 0.379%

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