When a dart is thrown at a circular target, consider the location of the landing

Question

When a dart is thrown at a circular target, consider the location of the landing point relative to the bull's eye. Let X be the angle in degrees measured from the horizontal, and assume that X is uniformly distributed on [0, 360]. Define Y to be the transformed variable Y = h(X) = (2/360)X-, so Y is the angle measured in radians and Y is between - and . Obtain E(Y) and Y by first obtaining E(X) and X, and then using the fact that h(X) is linear function of X.

I know the answer is approximately 1.814, but need to understand how to find this value.

Explanation / Answer

We are given that X is uniformly distributed on the interval [0, 360]. The general formulae for the mean and variance of a random variable U that is uniformly distributed on the interval [0, a] are E(U) = a/2 and Var(U) = (a^2)/12. In the case of X, a = 360. Therefore E(X) is E(X) = 360/2 = 180. Var(X) = 360 x360/12 = 10800. So sigma(X) = sqrt(10800) = 103.923. Y is a linear function of X. In general if Y = aX +b, where a and b are constants then E(Y) = a E(X) + b and Var(Y) = (a^2) Var(X). Hence E(Y) = (2 pi/360) x 180 - pi = pi -pi = 0. Var(Y) = (2 pi/360)^2 x 10800 = 3.290 Therefore, sigma(Y) = 1.814.

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