The driving distance for the top 100 golfers on the ABC tour is between 280.5 an

Question

The driving distance for the top 100 golfers on the ABC tour is between 280.5 and 315.3 yards. Assume that the driving distance for these golfers is uniformly distributed over this interval.

What is the probability the driving distance for one of these golfers is less than 290 yards (to 4 decimals)?

What is the probability the driving distance for one of these golfers is at least 300 yards (to 4 decimals)?

What is the probability the driving distance for one of these golfers is between 290 and 305 yards (to 4 decimals)?

Explanation / Answer

a.difference in interval = 315.3-280.5=34.8
p(<290)=(290-280.5)/34.8
=9.5/34.8
=0.2729

b. p(atleast 300 yards) =(315.3-300)/34.8
=15.3/34.8
=0.4396

c. p(between 290 and 305)
=305-290=15
=(305-290)/34.8=15.3/34.8=0.4310

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