Problem Breakdown:

We are dealing with a continuous random variable $X$ with the following properties:

• The probability density function $P(x) = 0.5$ for $x \in [10, 12]$.
• $P(x) = 0$ outside the interval $[10, 12]$.

Part 1: Finding $P(11)$

For a continuous random variable, the probability at any single point is always zero. This is a fundamental property of continuous distributions: the probability of observing a single specific value, such as $P(11)$, is always zero.

Thus, $P(11) = 0$

Part 2: Finding the Standard Deviation of $X$

To find the standard deviation, we need to determine both the mean $\mu$ and the variance $\sigma^2$ of $X$.

Step 1: Mean $\mu$

The mean for a continuous uniform distribution is given by the formula:

$\mu = \frac{a + b}{2}$ where $a = 10$ and $b = 12$ are the limits of the interval.

So, $\mu = \frac{10 + 12}{2} = 11$

Step 2: Variance $\sigma^2$

The variance for a uniform distribution on the interval $[a, b]$ is:

$\sigma^2 = \frac{(b - a)^2}{12}$ Substitute $a = 10$ and $b = 12$:

$\sigma^2 = \frac{(12 - 10)^2}{12} = \frac{4}{12} = \frac{1}{3}$

Step 3: Standard Deviation $\sigma$

The standard deviation is the square root of the variance:

$\sigma = \sqrt{\frac{1}{3}} \approx 0.577$

Final Answer:

• $P(11) = 0$
• The standard deviation of $X$ is approximately $0.577$.

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Would you like further details or have any questions about this solution?

Related Questions:

1. How would the standard deviation change if the interval for $X$ was different?
2. How do you calculate the mean and variance for a continuous uniform distribution?
3. Can the probability density function for a continuous random variable ever be negative?
4. How would the probability change if we were working with a discrete random variable instead?
5. What is the relationship between the area under a probability density function and the total probability?

Tip:

For continuous random variables, remember that the probability at a specific point is always zero, but the probability over an interval can be found by integrating the probability density function over that interval.