Question

1. the random variable x has cumulative distribution function: f(x) = { 0 x < -2 1/4 x + 1/2 -2 ≤ x ≤ 2 1 x > 2 the random variable y = 12x + 8 when x ≥ 0 and y = 0 if x is negative. find ey. a) 10 b) 0 c) 32 d) 8 e) 40 f) none of the above.

Explanation:

Step1: Find the probability density function \(f(x)\)

The cumulative distribution function \(F(x)\) is given by:

$$F(x)=
LATEXBLOCK0
$$

The probability density function \(f(x)\) is the derivative of \(F(x)\).
For \(x\in(-2,2)\), \(f(x)=\frac{d}{dx}(\frac{1}{4}x+\frac{1}{2})=\frac{1}{4}\). For \(x
otin(-2,2)\), \(f(x) = 0\).

Step2: Use the formula for \(E[Y]\)

We know that \(Y = 12X+8\) and \(E[Y]=E[12X + 8]=12E[X]+8\)
The formula for \(E[X]=\int_{-\infty}^{\infty}xf(x)dx\). Since \(f(x)=\frac{1}{4}\) for \(x\in[-2,2]\) and \(0\) otherwise.

$$E[X]=\int_{-2}^{2}x\cdot\frac{1}{4}dx$$

Using the integral formula \(\int_{a}^{b}x dx=\frac{x^{2}}{2}\big|_{a}^{b}\), we have:

$$E[X]=\frac{1}{4}\cdot\frac{x^{2}}{2}\big|_{-2}^{2}=\frac{1}{8}(2^{2}-(-2)^{2}) = 0$$

Step3: Calculate \(E[Y]\)

Substitute \(E[X]=0\) into \(E[Y]=12E[X]+8\)

$$E[Y]=12\times0 + 8=8$$

Answer:

D) 8