Continuous probability distributions

A continuous random variable X that can assume values between 0 and 1 has a density function given by:
\[ \begin{align} f(x) &= 2(1 - x) \quad 0 \le x \le 1 \\ &= 0 \qquad \text{elsewhere} \end{align}\] (a) Confirm this is a probability distribution.
(b) What is the probability of X taking a value less than 1/2?

Use the definition to confirm f is a probability distribution.

Now f(x) is \( \ge 0 \) for all values of x, and \[ \begin{align} \int_{-\infty}^{\infty}2(1 - x) dx &= \int_0^1 2(1 - x) dx \\ \end{align}\ \] because f is zero outside these limits. \[ \begin{align} \int_0^1 2(1 - x) dx &= \int_0^1 (2 - 2x) dx \\ &= \left[ 2x - x^2 \right]_0^1 \\ &= 1 \end{align}\ \] so f is a probability distribution.

To find the probability that x is less than 1/2, use the definition

\[ \begin{align} P(0 \le x \le 1) &= \int_0^{1/2} 2(1-x) dx \\ &= \left[ 2x - x^2 \right]_0^{1/2} \\ &= 1 - 1/4 \\ &= 0.75 \end{align}\ \]

You will gain most benefit if you work through each problem first and then come back and check your work against the solution.

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Now You Try:

A continuous random variable X that can assume values between 0 and 1 has a density function given by:
\[ \begin{align} f(x) &= 3x^2 \quad 0 \le x \le 1 \\ &= 0 \qquad \text{elsewhere} \end{align}\] (a) Confirm this is a probability distribution.
(b) What is the probability of X taking a value less than 1/2?