This problem involves a uniform distribution of a random variable $X$ over the interval $[0, 2]$.

Key Information:

1. The random variable $X$ is uniformly distributed over $[0, 2]$, meaning the probability density function (PDF) is constant on this interval.
2. The total probability over the entire interval $[0, 2]$ is 1, so the height of the PDF is given by: $f(x) = \frac{1}{\text{length of the interval}} = \frac{1}{2}$

Problem Statement:

We are asked to find the probability that $X$ falls between 0.7 and 1.66.

For a uniform distribution, the probability of $X$ being between two values $a$ and $b$ is the length of that interval $(b - a)$ multiplied by the height of the PDF $f(x)$: $P(a \leq X \leq b) = (b - a) \times f(x)$

Solution:

1. The length of the interval between 0.7 and 1.66 is: $1.66 - 0.7 = 0.96$
2. The height of the PDF is $f(x) = \frac{1}{2}$.

Thus, the probability is: $P(0.7 \leq X \leq 1.66) = 0.96 \times \frac{1}{2} = 0.48$

Therefore, the probability that a randomly selected $X$ is between 0.7 and 1.66 is 0.48.

Would you like any further clarification or details on this? Here are some related questions you might find useful:

1. How do you calculate the probability for a uniform distribution over a different interval?
2. What is the significance of the area under the probability density function (PDF)?
3. How would this problem change if the distribution were not uniform?
4. What is the cumulative distribution function (CDF) for a uniform distribution?
5. How does the probability change if the interval of $X$ is expanded?

Tip: For uniform distributions, the probability is proportional to the length of the interval, making it simple to compute as the area of a rectangle.