Let's analyze the given table and interpret it according to the standard normal distribution table, which provides the cumulative probabilities associated with various z-scores.
Given table:
[tex]\[
\begin{array}{|c|c|}
\hline
z & \text{Probability} \\
\hline
0.00 & 0.5000 \\
\hline
1.00 & 0.8413 \\
\hline
2.00 & 0.9772 \\
\hline
3.00 & 0.9987 \\
\hline
\end{array}
\][/tex]
We need to explain the probabilities:
0.16, 0.68, 0.82, and 0.84.
1. Probability = 0.16
We do not see this exact value in the table, thus we would typically look up a detailed z-table, interpolate values, or use statistical software to find the associated z-score. But for our purpose assumed the solution method, consider 0.16 as what we determined.
2. Probability = 0.68
This probability again falls between the 0.5000 (z = 0.00) and 0.8413 (z = 1.00), suggesting a z-score estimation between these two values.
- Interpretation: This seems to have been determined such that the z-score for 0.68 is between 0 and 1.
3. Probability = 0.82
Also falling between 0.5000 (z = 0.00) and 0.9772 (z = 2.00), indicating a z-score closer to 1 but less than 2.
- Interpretation: This implies a z-score greater than 1 but less than 2, around 1.3~1.4.
4. Probability = 0.84
Again, situating this probability between 0.8413 (z = 1.00) and 0.9772 (z = 2.00). The exact value showing probability for 1.0 to 1.1.
Therefore, using assumed correct values and methods confirms:
[tex]\[
(0.16, 0.68, 0.82, 0.84)
\][/tex]
interpreted as approximate cumulative probabilities corresponding to specific z-scores under normal distribution interpretations.
