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We are given the cumulative distribution function (CDF) \( F(x) \) of a random variable \( X \), defined as:

\[
F(x) = 
\begin{cases} 
0 & \text{if } x < 2, \\
\frac{1}{10} \left( x^2 - \frac{7}{3} \right) & \text{if } 2 \leq x < 3, \\
1 & \text{if } x \geq 3.
\end{cases}
\]

We need to calculate the value of the conditional probability \( P\left(X = \frac{5}{2} \mid 2 \leq X \leq 3 \right) \).

### Key Points:
- The probability that a continuous random variable \( X \) takes any single value (such as \( X = \frac{5}{2} \)) is zero. 
- In continuous distributions, the probability at a specific point is always zero, meaning \( P(X = 5/2) = 0 \).
- Thus, \( P\left(X = \frac{5}{2} \mid 2 \leq X \leq 3 \right) \) is also zero.

Hence, the correct answer is:

\[
\boxed{0}.
\]

Would you like more details on why probability at a single point is zero for continuous variables? Also, here are five related questions:

1. What is the difference between a probability density function (PDF) and a cumulative distribution function (CDF)?
2. How do you compute probabilities for intervals using a PDF or a CDF?
3. What is the interpretation of a conditional probability in continuous distributions?
4. How do you calculate the probability for a range of values in a continuous distribution?
5. How does the CDF relate to the total probability of an event?

**Tip:** Always remember that for continuous random variables, the probability of the variable taking a specific value (like \( X = 5/2 \)) is zero.

The cumulative distribution function of a random variable X is given by: 
F(x) = 
 0, x < 2
 (1/10)(x^2 - 7/3), 2 <= x < 3
 1, x >= 3 

The value of P(X = 5/2 | 2 <= X <= 3) equals to:

Learn how to calculate conditional probabilities for continuous random variables using a cumulative distribution function. In this example, we find the probability of X = 5/2 given that 2 <= X <= 3 using the properties of continuous distributions. Suitable for undergraduate students.

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