A fair die has six faces numbered from 1 to 6. The random variable $X$ is defined as twice the number appearing on the die, meaning:

$X = 2Y$

where $Y$ is the number shown on the die. Since $Y$ can take values $1,2,3,4,5,6$, the possible values of $X$ are:

$X \in \{2, 4, 6, 8, 10, 12\}$

Since the die is fair, the probability of each outcome is:

$P(Y = k) = \frac{1}{6}, \quad k = 1,2,3,4,5,6$

Since $X = 2Y$, the probability distribution of $X$ is:

$X$	2	4	6	8	10	12
$P(X)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

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(ii) Mean $E(X)$

The mean (expected value) is given by:

$E(X) = \sum X_i P(X_i)$

Substituting values:

$E(X) = 2 \times \frac{1}{6} + 4 \times \frac{1}{6} + 6 \times \frac{1}{6} + 8 \times \frac{1}{6} + 10 \times \frac{1}{6} + 12 \times \frac{1}{6}$

$= \frac{2+4+6+8+10+12}{6} = \frac{42}{6} = 7$

So, $E(X) = 7$.

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(iii) Variance $V(X)$

Variance is given by:

$V(X) = E(X^2) - [E(X)]^2$

First, compute $E(X^2)$:

$E(X^2) = \sum X_i^2 P(X_i)$

$= 2^2 \times \frac{1}{6} + 4^2 \times \frac{1}{6} + 6^2 \times \frac{1}{6} + 8^2 \times \frac{1}{6} + 10^2 \times \frac{1}{6} + 12^2 \times \frac{1}{6}$

$= \frac{4 + 16 + 36 + 64 + 100 + 144}{6}$

$= \frac{364}{6} = 60.67$

Now, compute the variance:

$V(X) = 60.67 - 7^2$

$= 60.67 - 49 = 11.67$

So, $V(X) = 11.67$.

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Final Answers

1. Probability Distribution of $X$:
   $P(X = x) = \frac{1}{6}, \quad x = 2, 4, 6, 8, 10, 12$

2. Mean $E(X)$:
   $7$

3. Variance $V(X)$:
   $11.67$

Would you like further explanations?

Related Questions

1. How do you compute the standard deviation from variance?
2. What is the cumulative distribution function (CDF) of $X$?
3. How would the probability distribution change if the die was biased?
4. Can we generalize this problem for a die with $n$ sides?
5. How does the transformation $X = 2Y$ affect variance compared to $Y$?

Tip:
The variance formula can be rewritten using $V(aX) = a^2 V(X)$, which simplifies calculations when a variable is scaled.