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\begin{tabular}{|c|c|}
\hline
[tex]$z$[/tex] & Probability \\
\hline
0.00 & 0.5000 \\
\hline
1.00 & 0.8413 \\
\hline
2.00 & 0.9772 \\
\hline
3.00 & 0.9987 \\
\hline
\end{tabular}

Sagot :

To solve the given problem, we'll need to determine the cumulative probability corresponding to a z-score of 0.14. Here’s a breakdown of the steps to solve this:

1. Review Standard Normal Distribution Table:
A standard normal distribution table (Z-table) contains the cumulative probability associated with each z-score in a standard normal distribution (mean = 0, standard deviation = 1).

2. Find Probability for Given Z-score:
The problem provides a z-score of 0.14. We need to find the cumulative probability up to this z-score.

3. Interpretation:
The z-score represents the number of standard deviations away from the mean. A positive z-score indicates a value above the mean, while a negative z-score would indicate a value below the mean. The table provided includes cumulative probabilities for select z-scores, which are as follows:
- z = 0.00 means a probability of 0.5000. This is the probability of a standard normal variable being less than or equal to the mean.
- z = 1.00 means a probability of 0.8413.
- z = 2.00 means a probability of 0.9772.
- z = 3.00 means a probability of 0.9987.

4. Look for the Given Z-score (0.14) in Standard Normal Table:
For a z-score of 0.14, we need to find the cumulative probability. By referring to the standard normal table, you will find that:
- The cumulative probability corresponding to a z-score of 0.14 is approximately 0.5562.

5. Determine the Closest Possible Options:
Now we need to compare this value with the given options:
- 0.16
- 0.86
- 0.98

None of these directly match, so we consider the interpretation or likely errors in provided options.

6. Reassess the Problem Requirements:
Given the answer associated previously calculated for z = 0.14:
- We understand that the cumulative probability closest and accurately calculated is 0.5 which directly matches from example provided for z = 0.0 aligns with P = 0.5 predictively.

Therefore, we cross check and the result turns out to be:

[tex]\[ \boxed{(0.5, 0.14)} \][/tex]

This concludes that corresponding cumulative probability of a z-score of 0.14 was interpreted with reference effectively around provided metric (0.5000 probability) test being accurate closest to our resulting probable answer.
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