In a lumberjack competition, a contestant is blindfolded and then spun around 9t

Question

In a lumberjack competition, a contestant is blindfolded and then spun around 9times. The contestant then tries to hit the target point in the middle of a horizontal log with an axe (while still blindfolded). The contestant receives 15 points if his hit is within 3 cm of the target, 10 points if his hit is between 3 cm and 10 cm off the target, 5 points if his hit is between 10 cm and 20 cm off the target, and zero points if his hit is beyond 20 cm away from the target (and someone may lose a finger!). Let Y record the position of the hit, so that Y = y > 0 corresponds to missing the target point to the right by y cm and Y = -y

Explanation / Answer

Given Y is the position of the hit, with Y=y>0 for the hit to the right and Y=-y <0 for the hit to the left. Also given that Y is normally distributed with mean 0 and variance 100, giving standard deviation of 10.

Receives 15 point for Y < 3 and Y>-3 (hit lying within 3cms of the target)

10 points for 3<=Y<=10 and -10>=Y>=-3

5 points for 10<Y<=20 and -20>=Y>-10

0 points for Y>20 and Y<-20.

Let the number of points be X. Expected number of points be E[X]. Let the probability of an event be P(event). Then E[X] should be obtained from conditional expectation as the number of points received at any hit depends on the position of hit, which is also a random variable.

E[X] = 15[P(Y < 3) + P(Y>-3)]+10[P(3<=Y<=10)+P(-10>=Y>=-3)]+5[P(10<Y<=20)+P(-20>=Y>-10)]+0.

The above equation can be explained as follows. Given that a particular hit is made, we know the number of points. Then the conditional expectation can be obtained by multiplying the probability of a particular hit with the corresponding number of points and taking sum over all possible hits.

Normal distribution is symmetric over mean, hence expectation can be simplified as

E[X] = 30P(Y<=3)+20P(3<=Y<=10)+10P(10<Y<=20)

Also to use the standard normal distribution table, we have to normalize values by mean and standard deviation (i.e Z = (Y-mean)/standard deviation) and the E[X] becomes

E[X] = 30P(Z<=0.3)+20P(0.3<=Z<=1)+10P(1<Z<=2)

= 30*0.6179+20(P(Z<=1)-P(Z<=0.3))+10(P(Z<=2)-P(Z<=1))

= 18.537+20(0.8413-0.6179)+10(0.9772-0.8413)

= 18.537+4.468+1.359=24.364

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