The problem involves a uniform distribution between $X = 0$ and $X = 2$, and you're asked to find $P(X \geq 1.82)$.

Step-by-Step Solution:

1. Understanding Uniform Distribution:

   • A uniform distribution means the probability is evenly distributed across the interval. The probability density function (PDF) of a uniform distribution between $a$ and $b$ is: $f(x) = \frac{1}{b - a}$ Here, $a = 0$ and $b = 2$, so: $f(x) = \frac{1}{2 - 0} = \frac{1}{2}$

2. Calculate the Required Probability:

   • You're asked to find $P(X \geq 1.82)$. This is the area under the PDF curve from $X = 1.82$ to $X = 2$, which is a simple proportion of the total interval.
   • The length of the interval from 1.82 to 2 is: $2 - 1.82 = 0.18$
   • The probability of being in this interval is: $P(X \geq 1.82) = \text{length of interval} \times \text{height of PDF} = 0.18 \times \frac{1}{2} = 0.09$

Thus, $P(X \geq 1.82) = 0.09$.

Would you like further details on this?

Here are 5 related questions for further practice:

1. What is $P(X \geq 1.5)$ for the same uniform distribution?
2. How would the probability change if the interval were $[0, 3]$ instead of $[0, 2]$?
3. What is the probability that $X$ is between 0.5 and 1.5?
4. How do you calculate the expected value of a uniform distribution?
5. What is the variance of $X$ in a uniform distribution between $a$ and $b$?

Tip: The area under the curve in a uniform distribution always represents probability!