Question

Miguel is a golfer, and he plays on the same course each week. The following table shows the probability distribution for his score on one particular hole, known as the Water Hole.

Score 3 4 5 6 7
Probability 0.15 0.40 0.25 0.15 0.05
Let the random variable X represent Miguel’s score on the Water Hole. In golf, lower scores are better.

(a) Suppose one of Miguel’s scores from the Water Hole is selected at random. What is the probability that Miguel’s score on the Water Hole is at most 5? Show your work.





The name of the Water Hole comes from the small lake that lies between the tee, where the ball is first hit, and the hole. Miguel has two approaches to hitting the ball from the tee, the short hit and the long hit. The short hit results in the ball landing before the lake. The values of X in the table are based on the short hit. The long hit, if successful, results in the ball traveling over the lake and landing on the other side.

A potential issue with the long hit is that the ball might land in the water, which is not a good outcome. Miguel thinks that if the long hit is successful, his expected value improves to 4.2. However, if the long hit fails and the ball lands in the water, his expected value would be worse and increases to 5.4.

(c) Suppose the probability of a successful long hit is 0.4. Which approach, the short hit or the long hit, is better in terms of improving the expected value of the score? Justify your answer.



(d) Let p represent the probability of a successful long hit. What values of p will make the long hit better than the short hit in terms of improving the expected value of the score? Explain your reasoning.

Answer 1

Answer:

  (a) 0.80

  (c) short hit is better

  (d) p > 0.7083

Step-by-step explanation:

(a) The probabilities for scores of 6 or higher are 0.15 and 0.05. Their total is 0.20, so the probability of a 5 or less is 1 -0.20 = 0.80.

__

(c) The expected score using the long hit approach is ...

  p(success)·(success score) +p(failure)·(failure score) = 0.4(4.2) +0.6(5.4) = 4.92

The expected score using the short hit approach is ...

  3·0.15 +4·0.40 +5·0.25 +6·0.15 +7·0.05 = 4.55

Miguel's expected score is lower using the short hit approach (4.55 vs 4.92).

__

(d) As we saw above, for the probability of a successful long hit of p, the expected value of the score is ...

  p(4.2) +(1 -p)(5.4) = 5.4 -1.2p . . . . . . expected long hit score

We want to find p such that this value is less than the short hit expected value:

  5.4 -1.2p < 4.55

  0.85 -1.2p < 0 . . . . . subtract 4.55

  0.7083 - p < 0 . . . . . divide by 1.2

  p > 0.7083 . . . . . . . . add p

If the probability of a successful long hit is p > 0.7083, then the expected long hit score is less than the expected short hit score.

Answer 2

The probability Miguel’s score on the Water Hole is most at 5 is 0.80 or 80%.

The expected value would be worse and increases to 4.55.

The short hit is better than the long hit.

The value of P>0.7 will make the long hit better than the short hit.

Given ,

Miguel’s score on the Water Hole. In golf, lower scores are better is represented by x.

• The probabilities for scores of 6 or higher are 0.15 and 0.05.

Their total is 0.20,

P(X ≤ 5) = 0.15 + 0.40 + 0.25

P(X ≤ 5) = 0.80

P(X ≤ 5) = 80%

So the probability of a 5 or less is 1 -0.20 = 0.80.

Therefore, there is 80% chance that Miguel’s score on the Water Hole is at most 5.

• Calculate and interpret the expected value of X.

The expected value of random variable X is given by

E(X) = X₃P₃ + X₄P₄ + X₅P₅ + X₆P₆ + X₇P₇

E(X) = 3*0.15 + 4*0.40 + 5*0.25 + 6*0.15 + 7*0.05

E(X) = 0.45 + 1.6 + 1.25 + 0.9 + 0.35

E(X) = 4.55

Therefore, the expected value of 4.55 represents the average score of Miguel.

• The probability of a successful long hit is given by

P(Successful) = 0.40

The probability of a unsuccessful long hit is given by

P(Unsuccessful) = 1 - P(Successful)

P(Unsuccessful) = 1 - 0.40

P(Unsuccessful) = 0.60

The expected value of successful long hit is given by

E(Successful) = 4.2

The expected value of Unsuccessful long hit is given by

E(Unsuccessful) = 5.4

So, the expected value of long hit is,

E(long hit) = P(Successful)*E(Successful) +P(Unsuccessful)*E(Unsuccessful)

E(long hit) = 0.40*4.2 + 0.60*5.4

E(long hit) = 1.68 + 3.24

E(long hit) = 4.92

Since the expected value of long hit is 4.92 which is greater than the value of short hit obtained in part b that is 4.55, therefore, it is better to go for short hit rather than for long hit.

• The expected value of  long hit is given by

E(long hit) = P(Successful)*E(Successful) + P(Unsuccessful)*E(Unsuccessful)

E(long hit) = P*4.2 + (1 - P)*5.4

We want to find the probability P that will make the long hit better than short hit

P*4.2 + (1 - P)*5.4 < 4.55

4.2P + 5.4 - 5.4P < 4.55

-1.2P + 5.4 < 4.55

-1.2P < -0.85

Multiply both sides by -1

1.2P > 0.85

P > 0.85/1.2

P > 0.7083

Therefore, the probability of long hit must be greater than 0.7083 that will make the long hit better than the short hit in terms of improving the expected value of the score.

For the more information about Probability distribution click the link given below.

https://brainly.in/question/21063009