- The expected value \(E[X]\) is calculated as \((4 \times 0.2) + (6 \times 0.4) + (8 \times 0.4) = 6.4\).
- The expected value of the squares \(E[X^2]\) is \((16 \times 0.2) + (36 \times 0.4) + (64 \times 0.4) = 43.2\).
- The variance \(\text{Var}(X) = E[X^2] - (E[X])^2 = 43.2 - (6.4)^2 = 2.24\).
- The standard deviation \(\sigma = \sqrt{2.24} \approx 1.5\), which matches the first option.
- Final answer: \(\boxed{1.5}\).
Explanation
- Analyze the given probability distribution
The problem provides the probability mass function (PMF) for a discrete random variable \(X\). The values taken by \(X\) are \(\{4, 6, 8\}\) with probabilities \(P(X=4) = \frac{1}{5}\), \(P(X=6) = \frac{2}{5}\), and \(P(X=8) = \frac{2}{5}\). Our goal is to find the standard deviation, \(\sigma\), of this random variable. We first verify the probabilities: \(\frac{1}{5} + \frac{2}{5} + \frac{2}{5} = \frac{5}{5} = 1\), which confirms it is a valid PMF.
- Calculate the expected value \(E[X]\)
The expected value \(E[X]\) is the weighted average of the possible values of \(X\). It is calculated as:
\[E[X] = \sum x \cdot P(X=x)\]
\[E[X] = (4 \cdot \frac{1}{5}) + (6 \cdot \frac{2}{5}) + (8 \cdot \frac{2}{5}) = \frac{4}{5} + \frac{12}{5} + \frac{16}{5} = \frac{32}{5} = 6.4\]
Thus, the expected value is \(6.4\).
- Calculate the expected value of the square, \(E[X^2]\)
Next, we calculate the expected value of the square, \(E[X^2]\):
\[E[X^2] = \sum x^2 \cdot P(X=x)\]
\[E[X^2] = (4^2 \cdot \frac{1}{5}) + (6^2 \cdot \frac{2}{5}) + (8^2 \cdot \frac{2}{5}) = (16 \cdot \frac{1}{5}) + (36 \cdot \frac{2}{5}) + (64 \cdot \frac{2}{5})\]
\[E[X^2] = \frac{16}{5} + \frac{72}{5} + \frac{128}{5} = \frac{216}{5} = 43.2\]
So, \(E[X^2] = 43.2\).
- Calculate the variance and the standard deviation
The variance, denoted by \(\text{Var}(X)\), is defined as the difference between the mean of the squares and the square of the mean:
\[\text{Var}(X) = E[X^2] - (E[X])^2\]
\[\text{Var}(X) = 43.2 - (6.4)^2 = 43.2 - 40.96 = 2.24\]
The standard deviation \(\sigma\) is the positive square root of the variance:
\[\sigma = \sqrt{\text{Var}(X)} = \sqrt{2.24}\]
Since \(\sqrt{2.24} \approx 1.49666...\), this is approximately \(1.5\). Comparing this to the given options (\(1.5\), \(5.79\), \(2\sqrt{6.4}\)), we conclude that \(1.5\) is the correct selection.
Examples
Standard deviation is essential in finance for measuring the volatility or risk of an investment. For instance, if an investor considers a portfolio with different potential return outcomes, the standard deviation helps quantify how much the actual returns might deviate from the expected average, allowing the investor to make informed decisions about their risk tolerance. In this case, calculating \(\sigma\) for a discrete probability distribution provides a clear understanding of the spread of potential outcomes.