I figured out C
310.6-300= 10.6
10.6(1/25.9)= .4093
Still trying to get the rest. Any tips would be appreciated
The driving distance for the top 100 golfers on the PGA tour is between 284.7and 310.6 yards (Golfweek, March 29, 2003). Assume that the driving distance for these golfers is uniformly distributed over this interval.
b.What is the probability the driving distance for one of these golfers is less than 290 yards (to 4 decimals)?
c.What is the probability the driving distance for one of these golfers is at least 300 yards (to 4 decimals)?
d.What is the probability the driving distance for one of these golfers is between 290 and 305 yards (to 4 decimals)?
e.How many of the 100 golfers drive the ball at least 290 yards?
I figured out C
310.6-300= 10.6
10.6(1/25.9)= .4093
Still trying to get the rest. Any tips would be appreciated
I figured out D
305-290=15
15(1/25.9)=.5792
Still trying to get B and E. Im having a hard time figuring these two out.
Well everytime I post I figure out another answer LOL. I figured out B now.
290-284.7=5.3
5.3(1/25.9)=.2046
Any suggestions on E
Okay I got the answer for E but i'm not sure if I did it right. Please give me feedback so I will know how to do it for the quiz next week.
e.How many of the 100 golfers drive the ball at least 290 yards?
310.6-290=20.6
100-20=80
80 is the answer but i'm not sure if it was done correctly.
I guess part e is asking for the expected number of golfers drive the ball at least 290 yards.
The total number of golfers drive the ball at least 290 yards is the sum of 100 i.i.d. Bernoulli random variables.
Since the expectation is linear, the expectation of the sum of the r.v.s is just the sum of the expectation of each individual r.v. Also the expectation of Bernoulli r.v. is just the probability, which is readily calculated. By using the i.i.d. property, the sum of expectation is just 100 times the individual expectation.
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