1. The problem is to understand why the probability is expressed as $P(Z \leq -2.33) = 0.0099$ instead of $P(Z > -2.33)$, especially when comparing it to a probability like $P(16.00 \leq X)$. 2. Here, $Z$ represents a standard normal variable, which is symmetric about zero. The value $-2.33$ is a z-score indicating how many standard deviations below the mean the value lies. 3. The notation $P(Z \leq -2.33)$ means the probability that $Z$ is less than or equal to $-2.33$, which corresponds to the left tail of the normal distribution. This is a cumulative probability from the far left up to $-2.33$. 4. The value $0.0099$ is the cumulative probability up to $-2.33$, meaning there is a 0.99% chance that $Z$ is less than or equal to $-2.33$. 5. When you see $P(16.00 \leq X)$, this is a probability of $X$ being greater than or equal to 16.00, which is a right-tail probability. The choice of less than or greater than depends on the problem context and which tail of the distribution you are interested in. 6. In summary, $P(Z \leq -2.33)$ is used because we are interested in the left tail probability up to $-2.33$, while $P(16.00 \leq X)$ is used when interested in the right tail starting from 16.00. The direction of inequality depends on which side of the distribution the event lies. 7. Important rule: For a standard normal variable $Z$, $P(Z \leq z)$ gives the cumulative probability from the left up to $z$, and $P(Z > z) = 1 - P(Z \leq z)$ gives the right tail probability beyond $z$. Final answer: The inequality direction depends on which tail of the distribution you want the probability for. $P(Z \leq -2.33)$ is the left tail probability, while $P(16.00 \leq X)$ is the right tail probability.