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Chapter 4: Problem 94

One professional golfer plays best on short-distance holes. Experience has shown that the numbers \(x\) of shots required for \(3-, 4-,\) and 5 -par holes have the probability distributions shown in the table: $$\begin{array}{lccccc}\text { Par-3 Holes } & \multicolumn{3}{c} {\text { Par-4 Holes }} & \multicolumn{2}{c} {\text { Par-5 Holes }} \\ \hline x & p(x) & x & p(x) & x & p(x) \\\\\hline 2 & .12 & 3 & .14 & 4 & .04 \\\3 & .80 & 4 & .80 & 5 & .80 \\\4 & .06 & 5 & .04 & 6 & .12 \\ 5 & .02 & 6 & .02 & 7 & .04\end{array}$$ What is the golfer's expected score on these holes? a. A par-3 hole b. A par- 4 hole c. A par-5 hole

Short Answer

Expert verified
Answer: The golfer's expected scores on each type of hole are: a. Par-3 hole: 2.98 b. Par-4 hole: 3.94 c. Par-5 hole: 5.16

Step by step solution

01

Expected score for a par-3 hole

To calculate the expected score for the par-3 hole, we use the given probabilities: \(E(X) = 2\times 0.12 + 3\times 0.80 + 4\times 0.06 + 5\times 0.02\) Now, we compute the result: \(E(X) = 0.24 + 2.4 + 0.24 + 0.10 = 2.98\) So, the expected score on a par-3 hole is 2.98.
02

Expected score for a par-4 hole

Similarly, for the par-4 hole, we use the probabilities: \(E(X) = 3\times 0.14 + 4\times 0.80 + 5\times 0.04 + 6\times 0.02\) Now, we compute the result: \(E(X) = 0.42 + 3.2 + 0.2 + 0.12 = 3.94\) So, the expected score on a par-4 hole is 3.94.
03

Expected score for a par-5 hole

Finally, for the par-5 hole, we use the probabilities: \(E(X) = 4\times 0.04 + 5\times 0.80 + 6\times 0.12 + 7\times 0.04\) Now, we compute the result: \(E(X) = 0.16 + 4 + 0.72 + 0.28 = 5.16\) So, the expected score on a par-5 hole is 5.16. In conclusion, the golfer's expected scores on each type of hole are: a. Par-3 hole: 2.98 b. Par-4 hole: 3.94 c. Par-5 hole: 5.16

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