The scores of students on a standardized test form a Normal distribution with a mean of 300 and a standard deviation of 40. Using the fact that the middle 50% of the data falls within 0.67 standard deviations from the mean, find the lower and upper quartile scores for this test.
A) 299.33 and 300.67
B) 273.2 and 326.8
C) 260 and 340
D) 150 and 450

Question

Quizplus · Accepted Answer

To find the lower and upper quartile, we need to find the z-scores that correspond to the 25th and 75th percentiles of the data. We know that the middle 50% of the data falls within 0.67 standard deviations from the mean. Therefore, the distance from the mean to the 25th percentile is -0.67 standard deviations, and the distance from the mean to the 75th percentile is +0.67 standard deviations.

Using a standard Normal distribution table (or a calculator), we can find that the z-score corresponding to the 25th percentile is -0.6745, and the z-score corresponding to the 75th percentile is +0.6745.

To find the actual lower and upper quartile scores, we can use the z-score formula:

z = (x - μ) / σ

where x is the score we want to find, μ is the mean (300), and σ is the standard deviation (40). Rearranging the formula, we get:

x = z * σ + μ

Plugging in the values, we get:

Lower quartile:

x = -0.6745 * 40 + 300 = 273.2

Upper quartile:

x = 0.6745 * 40 + 300 = 326.8

Therefore, the lower and upper quartile scores are 273.2 and 326.8, respectively, which corresponds to choice B.

The Scores of Students on a Standardized Test Form a Normal