. A archer is aiming at a target with the bullseye located at the origin. The ar

Question

. A archer is aiming at a target with the bullseye located at the origin. The archer’s shot lands at a distance from the bullseye with distribution N (0, 1). The archer scores 10 points if the shot is within 1 inch of the bullseye, 5 points if it lands between 1 and 3 inches from the bullseye, 3 points if it lands between 3 and 5 inches from the bullseye, and 0 otherwise. What is the expected score of the archer?

The correct answer is 8.40805, but I need a full explanation on how to do this problem.

Explanation / Answer

Here the archer;s shot is according to normal distribution.

so archer scores 10 points if the shot is within 1 inch of the bullseye.

So let say X is the archer's shot distance from origin.

Pr(-1 < X < 1) = (1) - (-1) = 0.841345 - 0.158655 = 0.6827

where is the normal standard cumulative distribution

Here archer scores 5 points if the shot is from 1 inch to 3 inch of the bullseys. So, here the value of X will vary from 1 to 3 .

Pr( 1 < l X l < 3) = Pr( -3 < X < -1) + Pr( 1 < X < 3) = NORM (-3 < X < -1) + NORM (1 < X < 3) = [(-1) - (-3)] + [ (3) - (1)]

= [0.158655 - 0.00135] + [0.99865 - 0.841345]

= 0.3146

so Now, 3 points if it lands between 3 and 5 inches from the bullseys.

Pr(3 < l X l < 5) = 1 - Pr(l X l < 3) = 1 - 0.6827 - 0.3146 = 0.0027

so NOw

the expected score of the archer = X Pr(X) = 10 * 0.6827 + 5 * 0.3146 + 3 * 0.0027 = 8.40805

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