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The ratings for “driving distance off the tee” forthe top 10 golfers on the PGA

ID: 2951156 • Letter: T

Question

The ratings for “driving distance off the tee” forthe top 10 golfers on the PGA Tour are as follows:


Prev Rank

Player

Value (yards)

1

1

ScottHend

318.90

2

2

TigerWoods

315.20

3

3

BrettWetterich

310.70

4

4

JohnDaly

308.50

5

6

ScottGutschewski

308.40

6

5

HankKuehne

308.30

7

10

Davis LoveIII

304.30




The Question:


How would you compute the probability, for any given drive onany given hole in any given tournament of the followingprobabilities for Tiger Woods, currently rated the Number Oneplayer in the world:


Note: Let D = the event “a given drive”


          I am NOTlooking for specific probability numbers – I want to knowhow


            you would compute them and what additional data you might needto


            make the computation using the methodology you suggest.


a) P(D = 310 yards) = ? How would you determine thisprobability? What assumptions did you make?


b) P(300 < D < 320) = ? How would youdetermine this probability? What

assumptions did you make?


c) P(300 < D < 320) = ? Howwould you determine this probability? What assumptions didyou make?

Rank

Prev Rank

Player

Value (yards)

1

1

ScottHend

318.90

2

2

TigerWoods

315.20

3

3

BrettWetterich

310.70

4

4

JohnDaly

308.50

5

6

ScottGutschewski

308.40

6

5

HankKuehne

308.30

7

10

Davis LoveIII

304.30

Explanation / Answer

one way would be to assume the distribution of the lengths of TigerWoods shots is a normal distribution . You would need the standard deviation of the distribution oflengths of tiger Woods shots and work with z tables. Sine the normal is continuous P(D=310)=0 and same reason band c would have the same answer. To work P(300
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