Let's carefully work through the given data to understand the solution step-by-step.
We are provided with a table that shows the cumulative probability corresponding to different [tex]\( z \)[/tex]-values:
[tex]\[
\begin{tabular}{|c|c|}
\hline
\(\mathbf{z}\) & \(\mathbf{P(Z \leq z)}\) \\
\hline
0.00 & 0.5000 \\
\hline
1.00 & 0.8413 \\
\hline
2.00 & 0.9772 \\
\hline
3.00 & 0.9987 \\
\hline
\end{tabular}
\][/tex]
The cumulative probability, [tex]\(P(Z \leq z)\)[/tex], is the probability that a standard normal random variable [tex]\(Z\)[/tex] is less than or equal to [tex]\(z\)[/tex].
From the table:
1. For [tex]\(z = 0.00\)[/tex], the cumulative probability [tex]\(P(Z \leq 0.00)\)[/tex] is 0.5000.
2. For [tex]\(z = 1.00\)[/tex], the cumulative probability [tex]\(P(Z \leq 1.00)\)[/tex] is 0.8413.
3. For [tex]\(z = 2.00\)[/tex], the cumulative probability [tex]\(P(Z \leq 2.00)\)[/tex] is 0.9772.
4. For [tex]\(z = 3.00\)[/tex], the cumulative probability [tex]\(P(Z \leq 3.00)\)[/tex] is 0.9987.
We are also given an intermediate probability of 0.16. This intermediate probability likely represents some fraction of the total probability distribution or may be a part of a calculation or condition, but it is not directly clear from the given information. It should be considered as an essential part of the problem, and it plays a specific role in the solution.
Summarizing:
- [tex]\(P(Z \leq 0.00) = 0.5000\)[/tex]
- [tex]\(P(Z \leq 1.00) = 0.8413\)[/tex]
- [tex]\(P(Z \leq 2.00) = 0.9772\)[/tex]
- [tex]\(P(Z \leq 3.00) = 0.9987\)[/tex]
- Intermediate probability = 0.16
Given this information and understanding the relationship between the z-values and their corresponding cumulative probabilities, these are the accurate values and should provide all needed insights for the problem at hand.
