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The expected value is essentially the "average" or "mean" value that a continuous random variable assumes. It's represented by \(E(X)\). For continuous random variables, it's determined using the formula:
\[E(X) = \int_{-\infty}^{\infty} x \cdot f(x) \, dx\]
This integral computes the weighted average across the range of \(X\) values, where each weight is given by the PDF. For our specific function \(f(x) = 6x(1-x)\) in the interval from 0 to 1, the expected value is calculated as:
\[E(X) = \int_{0}^{1} x \cdot 6x(1-x) \, dx\]
Through solving this integral, you can find the mean, which aids in understanding the central tendency of the distribution.

Variance measures the spread or dispersion of a set of values. For a continuous random variable \(X\), variance is calculated as:
\[Var(X) = E(X^2) - [E(X)]^2\]
Here, \(E(X^2)\) is the expected value of the square of \(X\), found by:
\[E(X^2) = \int_{-finity}^{finity} x^2 \cdot f(x) \, dx\]
For \(f(x) = 6x(1-x)\) within the range 0 to 1, it simplifies to:
\[E(X^2) = \int_{0}^{1} x^2 \cdot 6x(1-x) \, dx\]
The variance is critical for understanding how spread out the values of \(X\) are around the mean. A larger variance indicates more spread.

Standard deviation is a measure of how much the values of a distribution deviate from the mean on average. It is the square root of the variance:
\[SD(X) = \sqrt{Var(X)}\]
Standard deviation is easier to interpret compared to variance because it is in the same units as the variable, unlike variance which is in units squared. To find this, you determine the variance as seen in prior steps, and then compute its square root.
This value allows you to assess how tightly clustered the values of \(X\) are, making it simple to gauge the data dispersion about the mean. When asking for \(X\) within two standard deviations of the mean, you're effectively focusing on values that are close to the expected value in practical terms.