Final answer:
To find the time for 90% of test takers to complete a test normally distributed with a mean of 55 minutes and standard deviation of 6 minutes, use the Z-score formula to determine that approximately 62.68 minutes is sufficient. To find the proportion who need more than 70 minutes, calculate a Z-score of 2.5, which correlates to about 0.62% of students needing special arrangements.
Step-by-step explanation:
The question concerns the normal distribution of test completion times, with a mean of 55 minutes and a standard deviation of 6 minutes. To find the time needed to finish the test for 90% of test takers, we can use the Z-score table and look for the value that corresponds to the 90th percentile, which is approximately 1.28. We then apply the formula: X = μ + Z*σ, where X is the time for the 90th percentile, μ is the mean, Z is the Z-score, and σ is the standard deviation.
To calculate this, X = 55 + (1.28 * 6), which gives us X ≈ 62.68 minutes. Therefore, approximately 62.68 minutes is enough time to finish the test for 90% of the test takers.
To find the proportion of students who need more than 70 minutes for the test, and hence require special arrangements, we calculate the Z-score for 70 minutes: Z = (X - μ) / σ, which is (70 - 55) / 6 ≈ 2.5. Referring to the Z-score table, a Z-score of 2.5 corresponds to a cumulative proportion of about 0.9938. Therefore, the proportion of students expected to finish the test in less than 70 minutes is 0.9938, and the proportion of students who need special arrangements is 1 - 0.9938 ≈ 0.0062 or 0.62%.