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The Probability Density Function (PDF) gives us insight into where the values of a continuous random variable are most likely to occur. Imagine a curve that tells us the likelihood of different outcomes of a random event. The area under the curve between two points gives the probability that the variable falls within that interval, and this area must be equal to 1 across the entire range reflecting total certainty.
In our exercise's solution, we derived the PDF from the Cumulative Distribution Function (CDF) by differentiating it. Specifically, if the CDF is given by a function involving the variable, say for the range between 5 and 7 here, it was differentiated to find the PDF. Thus, the PDF for this problem was identified as:
\[ f(x) = \frac{1}{8}(3x^2 - 42x + 147) \]
This function describes the probability density across the variable's domain, here being 5 to 7. Notice how from the CDF we derive specific behavior of the variable within these bounds.

The Expected Value (often symbolized as \( E(X) \)) is like a weighted average for random variables, giving us a measure of the "central" tendency or the "mean" of our variable. It's calculated by integrating \( x \) times the PDF across the range of \( x \), which provides a balance point of the distribution.
The formula for the expected value for a continuous random variable \( X \) is:

• \[ E(X) = \int_{a}^{b} x f(x) \, dx \]

where \( f(x) \) is our PDF.
For our solved problem, this calculation involves determining \( E(X) \) over the interval from 5 to 7. This is done by evaluating:
\[ E(X) = \int_{5}^{7} x \left( \frac{1}{8} (3x^2 - 42x + 147) \right) dx \]
This expected value (what you "expect" the outcome to be in the long run if you repeated this random process many times) can be used to measure how the data is centered in this range.

Variance measures the spread of a random variable, telling us how much the values deviate from the Expected Value (or mean). Utilizing the formula:

• \[ \text{Var}(X) = E(X^2) - [E(X)]^2 \]

It ensures that we measure not just the "center" but how "scattered" the variables are around that center.
To find \( Var(X) \), first calculate \( E(X) \) and \( E(X^2) \):

• Calculate \( E(X^2) \) by evaluating the integral \( \int_{5}^{7} x^2 f(x) \, dx \)
• Subtract the square of \( E(X) \) from \( E(X^2) \)

This gives us a number representing variability. The larger the variance, the more spread out the data tends to be.

The median is a point at which half the data is above and half below. For continuous distributions, the median is the value \( x_m \) where the cumulative distribution function (CDF) equals 0.5.
In our context, the median of \( X \) is found by setting the CDF equation equal to 0.5 and solving for \( x_m \). This involves:

• Solving: \[ k(x_m^3 - 21x_m^2 + 147x_m - 335) = 0.5 \]
• Substituting the known \( k = \frac{1}{8} \)
• Re-arranging to simplify and solve the cubic equation \( x_m^3 - 21x_m^2 + 147x_m - 339 = 0 \)

The solution requires numerical methods as it's a non-trivial polynomial equation. The result reveals the median of the distribution, a solid representation of its "middle" point, fully partitioning the probability density into half above and half below.