Given a normally shaped distribution with a μ = 80 and a σ = 8, what scores are so extreme that they are achieved by only 2% of the population?
A) 63.52 and 96.48
B) 61.36 and 98.64
C) 55.36 and 104.64
D) 56.96 and 103.04

Question

Quizplus · Accepted Answer

Using a z-score table or calculator, we can find that the z-score associated with the 2nd percentile (2%) is approximately -2.05. We can then use the formula:

z = (x - μ) / σ

Rearranging, we can solve for x:

x = z * σ + μ

Substituting in the values of z, σ, and μ, we get:

x = -2.05 * 8 + 80 = 63.6

So, approximately 2% of the population will have a score below 63.6. To find the extreme scores achieved by only 2% of the population, we can subtract this value from the mean and double the result to get the range:

80 - 2(80 - 63.6) = 61.2

80 + 2(80 - 63.6) = 98.8

Therefore, the scores that are so extreme that they are achieved by only 2% of the population are 61.36 and 98.64, which corresponds to a z-score of -2.05 and +2.05, respectively.

Given a Normally Shaped Distribution with a μ = 80