Suppose that X is a random variable distributed according to the following function probability density: f(x)= 2(1 − x) for 0 ≤ x ≤ 1 f(x)= 0 otherwise. Since Y = 6X + 10, obtain variance of Y

To find the variance of Y, we need to first find the mean and variance of Y.

Step 1: Find the mean of Y.
The mean of Y can be found using the linearity property of expectation:
E[Y] = E[6X + 10] = 6E[X] + 10

E[X] = \int_{0}^{1} x \cdot 2(1 - x) \,dx = 2\int_{0}^{1} (x - x^2) \,dx
= 2\left[\frac{x^2}{2} - \frac{x^3}{3}\right] \bigg|_{0}^{1} = 2\left(\frac{1}{2} - \frac{1}{3}\right) = \frac{1}{3}

Therefore,
E[Y] = 6 \times \frac{1}{3} + 10 = 2 + 10 = 12

Step 2: Find the variance of Y.
The variance of Y can be found using the property: Var(aX + b) = a^2 Var(X), where a and b are constants.
Var(Y) = Var(6X + 10) = 6^2 Var(X) = 36 Var(X)

Var(X) = E[X^2] - (E[X])^2 = \int_{0}^{1} x^2 \cdot 2(1 - x) \,dx - \left(\frac{1}{3}\right)^2
= 2\int_{0}^{1} (x^2 - x^3)\, dx - \frac{1}{9} = 2\left[\frac{x^3}{3} - \frac{x^4}{4}\right] \bigg|_{0}^{1} - \frac{1}{9}
=2\left(\frac{1}{3}-\frac{1}{4}\right)-\frac{1}{9}=\frac{1}{6}-\frac{1}{9}=\frac{1}{18}

Therefore,
Var(Y)=36\times\frac{1}{18}=2