Question

In a lumberjack competition, a contestant is blindfolded and then spun around 9 times. The contestant then immediately tries to hit a single point (the target) in the middle of a horizontal log with an axe (while still blindfolded). The contestant receives

15 points if their swing is within 3cm of the target.
10 points if their swing is between 3cm and 10cm of the target.
5 points if their swing is between 10cm and 20cm of the target
zero points if their swing is further 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 < 0 corresponds to missing the target to the left by y cm. Assume that Y is normally distributed with mean mu = 0 and variance 100 cm^2. Find the expected number of points that the contestant wins.

Answer 1

9.364 is the expected number of points.

we can approximate this to 10 points if we want a whole number

Explanation:

We have these variables:

[0,5,10,15]

P(x= 0) = p(y>20)+p(y<-20) = 2p(y>20)

P(x=5) = p(-20<=y<=10)+p(10<=y<=20) = 2p(10<=y<=20)

P(x=10) = p(-10<=y<=-3)+p(3<=y<=10) = 2p(3<=y<=10)

P(x=15) = p(-3<=y<=3) = 2p(0<=y<=3)

Z = y/10

Therefore

P(x= 0) = 2(y>20)

= 2p(z>2) = 2(1-p<=2)

= 2(1-0.9772)

= 0.0456

P(x= 5)

= 2p(10<=y<=20)

= 2p(1<=z<=2)

= 2(0.9772-0.8413)

= 0.2718

P(x= 10)

= 2p(3<=y<=10)

= 2p(0.3<=z<=1)

= 2(0.8413-0.6179)

= 0.4468

P(x = 15)

= P(0.6179-0.3821)

= 0.2358

To get expected value of Y

0(0.0456)+5(0.2718)+10(0.4468)+15(0.2358)

= 1.359 + 4.468 + 3.537

= 9.364

E[Y] = 9.364