Did you know?
Did you know that in statistics, z-scores can help make calculations easier and more standardized? For example, when analyzing scores on a standardized test with a normal distribution (mean = 50 and standard deviation = 10), one can convert the test scores into z-scores to simplify calculations.
a) To determine the proportion of scores falling less than 65, we can convert 65 to a z-score using the formula z = (test score - mean) / sd. By doing so, we find that the z-score for 65 is (65-50)/10 = 1.5. Looking at a standard normal distribution table or using statistical software, we can find that the proportion of scores falling less than a z-score of 1.5 is approximately 0.9332. Therefore, about 93.32% of scores fall less than 65.
b) To find the proportion of scores falling between 40 and 65, we first convert these scores to z-scores. The z-score for 40 is (40-50)/10 = -1, and the z-score for 65 is 1.5 (as calculated in part a). Using the standard normal distribution table or software, we can find that the proportion of scores falling between a z-score of -1 and 1.5 is approximately 0.7745. This means that about 77.45% of scores fall between 40 and 65.
c) The 90th percentile represents the score below which 90% of the scores fall. By using the z-score formula, we can find the z-score corresponding to the 90th percentile. Letting the z-score be denoted by Z, we can solve the equation Z = (x - 50) / 10 for x. Rearranging the equation, we get x = 10Z + 50. Given that the 90th percentile corresponds to a z-score of 1.28 (obtained from the standard normal distribution table or software), plugging this value into the equation we find x to be 10(1.28) + 50 = 62.8. Therefore, the 90th percentile of scores on the test is approximately 62.8.
