**Problem: Random Variables and Their Distributions**Two independent random variables, \(X\) and \(Y\), are both uniformly distributed on \([0, 1]\). The random variable \(Z\) is defined by\[Z = (X - \mu_x)^2 + (Y - \mu_y)^2,\]where \(\mu_x = \mathbb{E}[X]\) and \(\mu_y = \mathbb{E}[Y]\).(a) Determine the expected value of \(Z\).(b) Determine the variance of \(Z\).

a) To determine the expected value (mean) of the random variable Z, we can use the properties of expected values and the given information about X and Y.First, let's find the expected values of X and Y:Since X and Y are uniformly distributed on [0,1], their expected values are the midpoints of the interval [0,1]:E[X] = μx = (0 + 1) / 2 = 1/2E[Y] = μy = (0 + 1) / 2 = 1/2Now, we can calculate the expected value of Z using the definition of Z:Z = (X - μx)2 + (Y - μy)2Now, let's calculate E[Z]:E[Z] = E[(X - μx)2 + (Y - μy)2]Using the properties of expected values, we can split this into two parts:E[Z] = E[(X - μx)2] + E[(Y - μy)2]Now, let's calculate each part separately:E[(X - μx)2]:E[(X - μx)2] is the variance of X:Var(X) = E[(X - μx)2] Since X is uniformly distributed on [0,1], its variance is (1/12): Var(X) = 1/12E[(Y - μy)2]:E[(Y - μy)2] is the variance of Y, which is also (1/12): Var(Y) = E[(Y - μy)2] = 1/12Now, add the two variances together to find the expected value of Z:E[Z] = Var(X) + Var(Y) = (1/12) + (1/12) = 1/6b) To determine the variance of the random variable Z, we can use the properties of variances and the given information about X and Y.Recall that we've already found the expected value of Z in part (a) as E[Z] = 1/6.Now, let's calculate the variance of Z:Var(Z) = E[Z2] - (E[Z])2First, we need to find E[Z2]:Z2 = (X - μx)2 + (Y - μy)2E[Z2] = E[(X - μx)2 + (Y - μy)2]Using the properties of expected values, we can split this into two parts:E[Z2] = E[(X - μx)2] + E[(Y - μy)2]E[(X - μx)2]:As we calculated in part (a), the variance of X is Var(X) = 1/12. We can use this value to find E[(X - μx)2]:E[(X - μx)2] = Var(X) + (E[X] - μx)2 E[(X - μx)2] = (1/12) + (1/2 - 1/2)2 E[(X - μx)2] = 1/12E[(Y - μy)2]:Similarly, the variance of Y is Var(Y) = 1/12.We can use this value to find E[(Y - μy)2]:E[(Y - μy)2] = Var(Y) + (E[Y] - μy)2 E[(Y - μy)2] = (1/12) + (1/2 - 1/2)2 E[(Y - μy)2] = 1/12Now, add the two values together to find E[Z^2]:E[Z2] = E[(X - μx)2] + E[(Y - μy)2] = 1/12 + 1/12 = 1/6Now, we can calculate Var(Z):Var(Z) = E[Z2] - (E[Z])2 = (1/6) - (1/6)2 = 1/6 - 1/36 = 5/36a) So, the expected value of Z is 1/6.b) So, the variance of Z is 5/36.