Finding the Median of a Continuous Random Variable

The question asks us to find the median of a continuous random variable X with a given probability density function (PDF).

The PDF is given as $f(x) = \frac{1}{8}$ for $-3 < x < 5$. This means the random variable X is uniformly distributed over the interval $(-3, 5)$. For values of x outside this interval, the PDF is zero.

Understanding the Median

For a continuous random variable, the median is the value 'm' such that the probability of the variable being less than or equal to 'm' is 0.5. Mathematically, this is expressed as:

$\int_{-\infty}^{m} f(x) dx = 0.5$

In this case, the PDF is non-zero only between -3 and 5. So, the integral needs to be evaluated from the start of the non-zero range up to 'm'. The median 'm' must lie within the interval $(-3, 5)$.

So, we need to solve:

$\int_{-3}^{m} \frac{1}{8} dx = 0.5$

Calculating the Median Step-by-Step

Let's evaluate the integral:

$\int_{-3}^{m} \frac{1}{8} dx = \frac{1}{8} \int_{-3}^{m} 1 dx$

The integral of 1 with respect to x is x.

$\frac{1}{8} [x]_{-3}^{m} = \frac{1}{8} (m - (-3))$

$\frac{1}{8} (m + 3)$

Now, we set this expression equal to 0.5:

$\frac{1}{8} (m + 3) = 0.5$

To solve for 'm', we can multiply both sides by 8:

$m + 3 = 0.5 \times 8$

$m + 3 = 4$

Subtract 3 from both sides:

$m = 4 - 3$

$m = 1$

The calculated median is 1.

We should check if this median lies within the given range of the distribution, which is $(-3, 5)$. Since 1 is indeed between -3 and 5, our result is valid.

Comparing with Options

Let's look at the given options:

• Option 1: 1
• Option 2: $\sqrt { \frac{64}{12}}$
• Option 3: 0
• Option 4: $\frac{64}{12}$

Our calculated median is 1, which matches Option 1.

Let's briefly check the other options:

• For m = 0 (Option 3): $\int_{-3}^{0} \frac{1}{8} dx = \frac{1}{8} [x]_{-3}^{0} = \frac{1}{8}(0 - (-3)) = \frac{3}{8} \ne 0.5$. So, 0 is not the median.
• Option 4: $\frac{64}{12} = \frac{16}{3} \approx 5.33$. This value is outside the domain of the PDF $(-3, 5)$, so it cannot be the median.
• Option 2: $\sqrt{\frac{64}{12}} = \sqrt{\frac{16}{3}} \approx \sqrt{5.33} \approx 2.31$. This value is within the domain, but we already found the median to be 1. Let's verify: $\int_{-3}^{\sqrt{16/3}} \frac{1}{8} dx = \frac{1}{8} [\sqrt{16/3} - (-3)] = \frac{1}{8} (\sqrt{16/3} + 3) = \frac{1}{8} (4/\sqrt{3} + 3) \approx \frac{1}{8} (2.31 + 3) = \frac{5.31}{8} \approx 0.66 \ne 0.5$.

Therefore, the correct median is 1.

Revision Table: Continuous Random Variable Median

Additional Information: Properties of Uniform Distribution

The continuous uniform distribution on the interval [a, b] has the PDF:

$f(x) = \begin{cases} \frac{1}{b-a} & \text{for } a \le x \le b \\ 0 & \text{otherwise} \end{cases}$

In this problem, the interval is $(-3, 5)$, so $a = -3$ and $b = 5$. The PDF is $f(x) = \frac{1}{5 - (-3)} = \frac{1}{8}$. This confirms the distribution is uniform.

For a continuous uniform distribution on $[a, b]$, the median is simply the midpoint of the interval.

Median $= \frac{a+b}{2}$

Using this formula for our problem with $a = -3$ and $b = 5$:

Median $= \frac{-3 + 5}{2} = \frac{2}{2} = 1$

This shortcut for uniform distributions confirms our result obtained by integrating the PDF.