A uniform distribution is defined by a constant probability density function (PDF) between two endpoints, $a$ and $b$. For a continuous uniform distribution, the probability that $x$ lies between two values $x_1$ and $x_2$ (where $a \leq x_1 < x_2 \leq b$) is calculated as the ratio of the interval length to the total length of the distribution.

Given the uniform distribution from $a = 3$ to $b = 26$, the PDF is:

$f(x) = \frac{1}{b - a} = \frac{1}{26 - 3} = \frac{1}{23}$

The probability that $x$ lies between two values $x_1$ and $x_2$ is given by:

$P(x_1 \leq x \leq x_2) = \frac{x_2 - x_1}{b - a}$

Let's calculate the probabilities for each part.

(a) Probability that $x$ lies between 4 and 18:

$P(4 \leq x \leq 18) = \frac{18 - 4}{26 - 3} = \frac{14}{23}$

(b) Probability that $x$ lies between 5 and 11:

$P(5 \leq x \leq 11) = \frac{11 - 5}{26 - 3} = \frac{6}{23}$

(c) Probability that $x$ lies between 6 and 15:

$P(6 \leq x \leq 15) = \frac{15 - 6}{26 - 3} = \frac{9}{23}$

(d) Probability that $x$ lies between 8 and 19:

$P(8 \leq x \leq 19) = \frac{19 - 8}{26 - 3} = \frac{11}{23}$

Summary of Results:

• (a) $P(4 \leq x \leq 18) = \frac{14}{23}$
• (b) $P(5 \leq x \leq 11) = \frac{6}{23}$
• (c) $P(6 \leq x \leq 15) = \frac{9}{23}$
• (d) $P(8 \leq x \leq 19) = \frac{11}{23}$

Would you like more details or explanations on these results?

Here are 5 related questions to expand your understanding:

1. How does the uniform distribution differ from a normal distribution?
2. What is the cumulative distribution function (CDF) for a uniform distribution?
3. How do you calculate the expected value for a uniform distribution?
4. What happens if the interval $[a, b]$ changes in a uniform distribution?
5. How can uniform distribution be applied in real-life scenarios like random number generation?

Tip: When dealing with probabilities in uniform distributions, remember that the area under the curve is always evenly distributed, so calculating probabilities is just about finding the proportion of the interval.