Given that the random variable $X$ is uniformly distributed between 4 and 12, with a height of $\frac{1}{8}$, the probability that $X$ is at most 7 can be found by calculating the area under the probability density function (PDF) from 4 to 7.

Step 1: Length of the interval

The uniform distribution is defined between 4 and 12. The total length of the interval is: $12 - 4 = 8$

Step 2: PDF of uniform distribution

The height of the uniform distribution is given as $\frac{1}{8}$. This means that the probability density is uniform (constant) over the interval [4, 12].

Step 3: Calculating the area for $X \leq 7$

The probability that $X$ is at most 7 is the area of the rectangle under the PDF from 4 to 7. The length of this subinterval is: $7 - 4 = 3$ The area under the PDF is the length of this interval multiplied by the height of the uniform distribution, which is $\frac{1}{8}$: $\text{Area} = (7 - 4) \times \frac{1}{8} = 3 \times \frac{1}{8} = \frac{3}{8} = 0.375$

Final Answer:

$P(X \leq 7) = 0.375$ Thus, the probability that $X$ is at most 7 is $0.375$ (or $\frac{3}{8}$).

Would you like further details or have any questions?

Here are five related questions:

1. What is the probability that $X$ is between 5 and 10?
2. How would the probability change if $X$ was uniformly distributed between 3 and 15?
3. What is the cumulative distribution function (CDF) for $X$ in this uniform distribution?
4. Can you calculate the mean and variance of a uniform distribution?
5. How does a uniform distribution compare to a normal distribution?

Tip: For uniform distributions, the probability is always proportional to the length of the interval of interest.