A fair six-sided die is rolled, with X being the number on the uppermost face. The variance of X is:
The question asks for the variance of the number shown on the uppermost face when a fair six-sided die is rolled. This involves understanding probability distributions and how to calculate variance for a discrete random variable.
Let \(X\) be the random variable representing the number on the uppermost face when a fair six-sided die is rolled. The possible outcomes for \(X\) are the integers from 1 to 6.
Since the die is fair, each outcome has an equal probability. This is a discrete uniform distribution. The probability mass function is:
We can represent the distribution in a table:
| Outcome (\(x\)) | Probability (\(P(X=x)\)) |
|---|---|
| 1 | \(\dfrac{1}{6}\) |
| 2 | \(\dfrac{1}{6}\) |
| 3 | \(\dfrac{1}{6}\) |
| 4 | \(\dfrac{1}{6}\) |
| 5 | \(\dfrac{1}{6}\) |
| 6 | \(\dfrac{1}{6}\) |
The expected value, or mean, of a discrete random variable \(X\) is given by the formula:
$$E[X] = \sum x \cdot P(X=x)$$
For the fair six-sided die:
$$E[X] = \left(1 \cdot \dfrac{1}{6}\right) + \left(2 \cdot \dfrac{1}{6}\right) + \left(3 \cdot \dfrac{1}{6}\right) + \left(4 \cdot \dfrac{1}{6}\right) + \left(5 \cdot \dfrac{1}{6}\right) + \left(6 \cdot \dfrac{1}{6}\right)$$
$$E[X] = \dfrac{1}{6} (1 + 2 + 3 + 4 + 5 + 6) = \dfrac{1}{6} (21) = \dfrac{21}{6} = \dfrac{7}{2} = 3.5$$
The expected value of rolling a fair six-sided die is 3.5.
The variance of a random variable \(X\) measures how spread out the possible outcomes are from the expected value. The formula for variance is:
$$Var(X) = E[X^2] - (E[X])^2$$
First, we need to calculate \(E[X^2]\). This is the expected value of the square of the random variable:
$$E[X^2] = \sum x^2 \cdot P(X=x)$$
For the fair six-sided die:
$$E[X^2] = \left(1^2 \cdot \dfrac{1}{6}\right) + \left(2^2 \cdot \dfrac{1}{6}\right) + \left(3^2 \cdot \dfrac{1}{6}\right) + \left(4^2 \cdot \dfrac{1}{6}\right) + \left(5^2 \cdot \dfrac{1}{6}\right) + \left(6^2 \cdot \dfrac{1}{6}\right)$$
$$E[X^2] = \dfrac{1}{6} (1 + 4 + 9 + 16 + 25 + 36) = \dfrac{1}{6} (91) = \dfrac{91}{6}$$
Now, we can calculate the variance using the formula \(Var(X) = E[X^2] - (E[X])^2\):
$$Var(X) = \dfrac{91}{6} - \left(\dfrac{7}{2}\right)^2$$
$$Var(X) = \dfrac{91}{6} - \dfrac{49}{4}$$
To subtract these fractions, we find a common denominator, which is 12:
$$Var(X) = \dfrac{91 \cdot 2}{6 \cdot 2} - \dfrac{49 \cdot 3}{4 \cdot 3}$$
$$Var(X) = \dfrac{182}{12} - \dfrac{147}{12}$$
$$Var(X) = \dfrac{182 - 147}{12}$$
$$Var(X) = \dfrac{35}{12}$$
The variance of the number on the uppermost face of a fair six-sided die is \(\dfrac{35}{12}\).
Let's review the steps taken to calculate the variance:
The calculation is consistent and follows the standard method for finding the variance of a discrete probability distribution.
Based on the calculations, the variance of the number on the uppermost face when a fair six-sided die is rolled is \(\dfrac{35}{12}\).
| Concept | Definition/Formula | Application to Die Roll |
|---|---|---|
| Random Variable (\(X\)) | A variable whose value is a numerical outcome of a random phenomenon. | Number shown on the die (1 to 6). |
| Probability Distribution | Specifies the probabilities for each possible outcome of a random variable. | Discrete uniform distribution (\(P(X=x) = 1/6\)). |
| Expected Value (\(E[X]\)) | The average or mean of a random variable. \(E[X] = \sum x P(X=x)\). | \(\dfrac{7}{2}\) or 3.5. |
| Variance (\(Var(X)\)) | A measure of how spread out the outcomes are from the expected value. \(Var(X) = E[X^2] - (E[X])^2\). | \(\dfrac{35}{12}\). |
Variance is an important measure of dispersion in statistics. A higher variance indicates that the data points are more spread out from the mean, while a lower variance indicates they are clustered closer to the mean.
The standard deviation is the square root of the variance. It is often used because it is in the same units as the random variable itself, making it easier to interpret than variance.
For this problem, the standard deviation would be \(\sqrt{\dfrac{35}{12}}\).
Understanding expected value and variance is fundamental in probability and statistics for analyzing the properties of random variables and probability distributions.
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