Calculating Probability for Continuous Random Variable

The problem asks for the probability that a continuous random variable $X$, with a given probability density function (PDF), falls within a specific range. The PDF is defined as:

$f(x) = \begin{cases} 0.25x & \text{for } 1 \le x \le 3 \\ 0 & \text{otherwise} \end{cases}$

We need to find the probability $P(1 \le X \le \sqrt{5})$.

Probability Calculation using Integration

To find the probability for a continuous random variable, we integrate the PDF over the desired interval. The interval of interest is from $1$ to $\sqrt{5}$. Since $\sqrt{5}$ (approximately 2.236) is within the defined range of $f(x)$ (1 to 3), the calculation is straightforward:

$P(1 \le X \le \sqrt{5}) = \int_{1}^{\sqrt{5}} f(x) \, dx $

Substitute the function $f(x) = 0.25x$ into the integral:

$P(1 \le X \le \sqrt{5}) = \int_{1}^{\sqrt{5}} 0.25x \, dx $

Evaluating the Integral

Factor out the constant $0.25$ and integrate $x$ with respect to $x$:

$= 0.25 \int_{1}^{\sqrt{5}} x \, dx $
$= 0.25 \left[ \frac{x^2}{2} \right]_{1}^{\sqrt{5}} $

Now, apply the limits of integration (upper limit $\sqrt{5}$ and lower limit $1$):

$= 0.25 \left( \frac{(\sqrt{5})^2}{2} - \frac{1^2}{2} \right) $
$= 0.25 \left( \frac{5}{2} - \frac{1}{2} \right) $
$= 0.25 \left( \frac{4}{2} \right) $
$= 0.25 \times 2 $
$= 0.5 $

The calculated probability is $0.5$. This value lies between $0.5$ and $0.5$, confirming the provided answer statement.