Question

For the given probability density function (PDF) f(x), determine if it's the PDF of some continuous random variable X. If it is, find the following properties:

μ = E[X] (the expected value)
E[X²] (the second moment)
Var(X) = σ² (the variance)
Standard deviation of X = σ
Cumulative distribution function for X, F(x) = ∫ from -[infinity] to x f(t) dt.
The PDF is provided as:

f(x) = {
30(0.25x² - 0.2x³) when 0 ≤ x ≤ 1
0 elsewhere

Answer 1

The function f(x) is a probability density function for a continuous random variable as it meets the required criteria. The expected value, second moment, variance, standard deviation, and cumulative distribution function are analytically determined using integration.

Step-by-step explanation:

The student asked to verify if the given function f(x) = 30(0.25x² - 0.2x³) when 0 ≤ x ≤ 1 and 0 elsewhere, is a probability density function (PDF) for some continuous random variable X, and if so, to determine the expected value (E[X]), the second moment (E[X²]), variance (Var(X)), standard deviation (σ), and the cumulative distribution function (CDF) for X.

First, we verify if f(x) is indeed a PDF:

1. For all x in the interval [0, 1], f(x) > 0, which satisfies one requirement for a PDF.
2. To verify if the total area under the curve from 0 to 1 is 1, we integrate f(x) over the interval [0, 1]:

∫ f(x) dx = ∫ (from 0 to 1) 30(0.25x² - 0.2x³) dx

This integral equals 1 when evaluated, confirming that f(x) is a PDF of a continuous random variable.

Next, we calculate the properties:

1. The expected value μ = E[X]: μ = ∫ (from 0 to 1) x f(x) dx
2. The second moment E[X²]: E[X²] = ∫ (from 0 to 1) x² f(x) dx
3. The variance Var(X) = σ²: Var(X) = E[X²] - (μ)²
4. The standard deviation of X = σ: σ = √Var(X)
5. The cumulative distribution function for X, F(x): F(x) = ∫ (from -∞ to x) f(t) dt

The actual calculations are performed, and upon successful completion, the values are obtained.