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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.

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.

Respuesta :

skyluke89

a) 4.55

b) Short hit approach

Step-by-step explanation:

a)

The table giving the score and the probability of each value of the score is:

Score: 3 4 5 6 7

Probability: 0.15 0.40 0.25 0.15 0.05

We have that:

X = Miguel's score on the Water Hole

For a certain variable X which can take values [tex]x_i[/tex] with relative probabilities [tex]p_i[/tex], the expected value of the variable is:

[tex]E(X)=\sum x_i p_i[/tex]

where:

[tex]x_i[/tex] are the possible values of variable X

[tex]p_i[/tex] is the probability of each value: [tex]p_i = p(X=x_i)[/tex]

By applying the formula here, we can find the expected value of MIguel's score:

[tex]E(X)=3\cdot 0.15 + 4\cdot 0.40 + 5\cdot 0.25 + 6\cdot 0.15 + 7\cdot 0.05=4.55[/tex]

2)

The expected value of X found in part 1) is based on the short hit approach; in that case, the expected value of X was

[tex]E(X)=4.55[/tex]

In this part we want to study the long hit approach instead. We have:

- If the long hit is successfull, the expected value of X is

[tex]E(X)=4.2[/tex]

- If the long hit fails, the expected value of X is

[tex]E(X)=5.4[/tex]

The probability of a successfull long hit is:

[tex]p(S)=0.4[/tex]

While the probabilty for the long hit to be unsuccessfull is:

[tex]p(U)=1-p(S)=1-0.4=0.6[/tex]

So, the expected value of X with the long hit approach is:

[tex]E(X)=p(S)\cdot 4.2 + p(U)\cdot 5.4 = 0.4\cdot 4.2 + 0.6\cdot 5.4 =4.92[/tex]

If we compare this value with the expected value of X of part 1, which was

[tex]E(X)=4.55[/tex]

We see that the expected value for X is lower (=better) for the short hit approach, so Miguel should choose the short hit approach.

fichoh

Using the principle of discrete probability, the expected value of short hit is ;

  • E(X) = 4.55
  • short hit is better than long hit in improving expected score.

The discrete probability distribution is given thus :

  • Score:____ 3___ 4__ 5___ 6__ 7
  • Probability: 0.15 0.40 0.25 0.15 0.05

The expected value of the experiment :

  • E(X) = Σ[(X) × p(X)]

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

E(X) = 4.55

For a long hit :

  • P(success) = 0.4
  • P(failure) = 1 - 0.4 = 0.6

Expected value of Long hit :

E(long hit) × p(success) + E(long hit) × p(failure)

(4.2 × 0.4) + (5.4× 0.6)

1.68 + 3.24 = 4.92

The expected value of short hit is lower, hence, better than that of long hit.

Learn more : https://brainly.com/question/15521567

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