Final answer:
The z-score associated with an 88% confidence interval is approximately 1.55, as it captures the middle 88% of the distribution, leaving 6% in each tail.
Step-by-step explanation:
The z-score that corresponds with an 88% confidence interval can be determined by looking at the z-table and finding the z-score that leaves 6% in each tail (since 100% - 88% = 12%, and 12%/2 = 6%). The z-score that has 94% of the distribution to the left (or 0.06 to the right) is approximately 1.55. This is because the z-table provides the cumulative area from the far left of the distribution up to a certain z-score, and you want to capture the central 88%. This concept applies to nearly any other confidence level; for a confidence interval, you always look for the z-score that leaves half of the remaining percentage in each tail of the distribution.
The appropriate z-score for an 88% confidence interval is approximately 1.174.
To find the z-score, we can use a standard normal distribution table or calculator. The formula for the z-score is:
z = invNorm((1 + confidence level) / 2)
Substituting 88% as the confidence level, we get:
z = invNorm((1 + 0.88) / 2) = invNorm(0.94) = 1.174