Cumulative distributions

A continuous random variable X that can assume values between 0 and 1 has a cumulative distribution given by:
\[ \begin{align} F(x) &= \frac{x^2+4x}{5} \quad 0 \le x \le 1 \\ &= 0 \quad x \lt 0 \\ &= 1 \quad x \gt 1 \end{align}\] (a) Find the corresponding probability density function.
(b) What is the probability of X taking a value less than 0.75?

The density function is the derivative of the cumulative distribution.

\[ \begin{align} \frac{d}{dx} \frac{x^2+4x}{5} &= \frac{2}{5}(x+2) \end{align}\ \] so \[ \begin{align} f(x) &= \frac{2}{5}(x+2) \end{align}\ \]

To find the probability that x is less than 0.75, use the definition of F

\[ \begin{align} P(0 \lt X \le 0.75) &= F(0.75) - F(0) \\ &= \frac{1}{5}\left[(\frac{3}{4})^2 + 4\frac{3}{4}\right] - 0 \\ &= 0.71 \end{align}\ \]

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 cumulative distribution given by:
\[ \begin{align} F(x) &= 4x^3 - 3x^4 \quad 0 \le x \le 1 \\ &= 0 \quad x \lt 0 \\ &= 1 \quad x \gt 1 \end{align}\] (a) Find the corresponding probability density function.
(b) What is the probability of X taking a value less than 0.5?