91Ó°ÊÓ

Since the amount of money spent by a customer is uniformly distributed over \((0, 100)\), we can calculate the expected value (mean) and variance using the following formulas for a uniform distribution: Mean: \(E[X] = \frac{a + b}{2}\) Variance: \(Var(X) = \frac{(b - a)^2}{12}\) where \(X\) is the random variable representing the amount spent by a customer, and \(a\) and \(b\) are the lower and upper bounds of the distribution, respectively. In this case, \(a = 0\) and \(b = 100\). Mean: \(E[X] = \frac{0 + 100}{2} = 50\) Variance: \(Var(X) = \frac{(100 - 0)^2}{12} = \frac{10000}{12} = 833.33\)

Given that the number of customers entering a store on a given day follows a Poisson distribution with mean \(\lambda = 10\), we can use the properties of the Poisson distribution and the expected value and variance calculated in Step 1 to find the mean and variance of the amount of money that the store takes in on a given day. Mean: \(E[Y] = \lambda E[X]\) Variance: \(Var(Y) = \lambda (Var(X) + E[X]^2) - \lambda^2 E[X]^2\) where \(Y\) is the random variable representing the amount of money that the store takes in on a given day. Mean: \(E[Y] = 10 \times 50 = 500\) Variance: \(Var(Y) = 10 (833.33 + 50^2) - 10^2 \times 50^2 = 10 (833.33 + 2500) - 10^2 \times 2500 = 10 \times 3333.33 - 250000 = 33333.3 - 250000 = 83,333.3\) Therefore, the mean amount of money that the store takes in on a given day is \(500, and the variance is \)83,333.3