The problem gives a uniformly distributed random variable $X$, denoted as:

$X \sim U[30, 180]$

This indicates that $X$ is uniformly distributed between 30 and 180. We are asked to find the probability that a random value of $X$ lies between 70 and 130.

Step-by-step Solution:

1. Uniform Distribution Formula: The probability density function (PDF) of a uniform distribution on the interval $[a, b]$ is given by:

   $f(x) = \frac{1}{b - a} \quad \text{for } a \leq x \leq b$

   In this case, $a = 30$ and $b = 180$. So the PDF is:

   $f(x) = \frac{1}{180 - 30} = \frac{1}{150}$

2. Probability Calculation: The probability that $X$ lies between 70 and 130 is the ratio of the length of the interval $[70, 130]$ to the length of the entire interval $[30, 180]$:

   $P(70 \leq X \leq 130) = \frac{130 - 70}{180 - 30} = \frac{60}{150} = 0.4$

Thus, the probability that a random value of $X$ is between 70 and 130 is 0.4 or 40%.

Would you like more details on the steps involved? Here are 5 related questions to deepen your understanding:

1. How is the probability density function (PDF) for a uniform distribution derived?
2. What is the difference between a uniform distribution and other probability distributions like normal or exponential?
3. How do you calculate the mean and variance of a uniform distribution?
4. Can the uniform distribution be used to model real-world scenarios, and if so, what examples fit this distribution?
5. What is the probability of $X$ lying between other intervals, such as 50 and 120?

Tip: In uniform distributions, probabilities over intervals can be quickly calculated by comparing the length of the desired interval to the total length of the distribution's range.