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 a linear function of ?X?.

Solution 27E Step1: We have let X be the angle in degrees measured from the horizontal and assume that X is 2 uniformly distributed on [0, 360].define Y to be the transformed variable Y = h(x)= ( 360)X - So Y is the angle measured in radian and Y is between and + . We need to obtain E(Y) and by yhe first obtaining E(X) and , and xhen using the fact that h(x) is a linear function of X. Step2: Let us assume that f(x) = k Now, 360 fX(x) = f(x)dx=1 0 360 = kdx=1 0 Integrate above equation with respect to x then we get = k [x] 360 0 = k [360 - 0] = 360k k = 1 360 Therefore, the value of k is 0.0027. Step3: Consider, 360 E(X) = f (X) = xf(x)dx=1 0 360 = xkdx=1 0 Integrate above equation with respect to x then we get x2 360 = k [ 2 0 (360) 02 = k [ 2 2 ] = k [ 129600 ] 2 2 = 1 [129600 ] 129600 2 = 720 = 180. Therefore, the mean is E(X) = 180. Similarly, 360 E(X ) = f (x) = x f(x)dx=1 X 0 360 2 = x kdx=1 0 Integrate above equation with respect to x then we get x3 360 = k [ ]3 0 (360) 03 = k [ 3 3 ] = k [ 46656000 ] 3 2 = 1 [46656000 ] 46656000 3 = 1080 = 43200. Therefore, the mean is E(X ) = 43200. 2 2 Var(X) = E(X ) - [E(X)] = 43200 - [180] 2 = 43200 - 32400 = 10800. Step4: Let Y = ( 2 )x - 360 We know that = 180 Now, E(Y) = ( 2(180))180 - 180 360 = 180 - 180 = 0 Therefore, the mean of Y is 0. 2 2 Var(Y) = (360) var(X) 2 = (129600) 10800 = ( 2 ) 10800 32400 10800 = 32400 = 3 Therefore, the variance of Y 3 .