In this problem, the diameter of the aerosol particles is a continuous random variable that is uniformly distributed between 2 and 6 nanometers. This means that all values between 2 and 6 are equally likely.

In this situation, our Random Variable X is a Uniform Distribution i.e. \( X \sim \text{Uniform}(a, b) \) where a is 2 and b is 6.

So for the random variable \( X \sim \text{Uniform}(2, 6) \) , we have the density function (PDF) defined as

\( f(x) = \begin{cases} 0 \quad \quad \quad \text{ for x } \le \text{a or x } \ge \text{b} \\ \dfrac{1}{b-a} \quad \text{for} \quad \text{a} \leq \text{x} \leq \text{b} \end{cases} \)

= \( f(x) = \begin{cases} 0 \quad \quad \quad \text{ for x } \le \text{2 or x } \ge \text{6} \\ \dfrac{1}{6-2} \quad \text{for} \quad \text{2} \leq \text{x} \leq \text{6} \end{cases} \)

= \( f(x) = \begin{cases} 0 \quad \quad \quad \text{ for x } \le \text{2 or x } \ge \text{6} \\ \dfrac{1}{4} \quad \text{for} \quad \text{2} \leq \text{x} \leq \text{6} \end{cases} \)

And the CDF is given as

\( F(x) = P(x \leq X) = \int_{-\infty}^t f(t)dt \)

We have to find the probability that a randomly measured particle has a diameter greater than 3 nanometers. This means that we have to the probabilities of particles whose diameter is between 3 to 6.

Thus, \( \text{required } F(x) = P(x \geq 3) = \int_{3}^6 f(t)dt \)

\( = \operatorname{\Huge\int}_{3}^6 \dfrac{1}{4} dt \)

\( = \operatorname{\Huge\int}_{3}^6 \dfrac{1}{4} dt \)

The probability that a randomly measured particle has a diameter greater than 3 nanometers is the ratio of the number of particles with diameters greater than 3 to the total number of particles.

\( = \dfrac{1}{4} \cdot{t} \operatorname{\Huge\vert}_{3}^{6} \)

\( = \dfrac{1}{4} \cdot{(6 – 3)} \)

\( = \dfrac{3}{4} \)