If a population of scores is normally distributed, has a mean of 45 and a standard deviation of 6, the most extreme 5% of the scores lie beyond the score(s) of _________.
A) 35.13
B) 45.99
C) 56.76 and 33.24
D) 45.99 and 35.13

Question

Quizplus · Accepted Answer

In a normal distribution, the most extreme 5% of scores (2.5% in each tail) are found by calculating the z-scores corresponding to these percentiles and then converting these z-scores back to the original scale using the mean and standard deviation. The z-score for 2.5% in the lower tail is approximately -1.96, and for the upper tail (97.5%), it is +1.96. Using the formula for converting z-scores to the original scale (X = μ + Zσ), where μ is the mean and σ is the standard deviation, we get the scores for the lower and upper bounds as follows: Lower bound = 45 + (-1.96)(6) ≈ 33.24, and Upper bound = 45 + (1.96)(6) ≈ 56.76.

If a Population of Scores Is Normally Distributed, Has a Mean