In an English class last semester, Foofy earned a 76 1X 5 85, SX 5 10). Her
friend, Bubbles, in a different class, earned a 60 1X 5 50, SX 5 4). Should Foofy be bragging about how much better she did? Why?
I don't understand your notation.
However, to answer your question, convert the raw scores to standard scores, like the Z score.
Z = (score-mean)/SD
Which Z score, if any, is higher?
To determine whether Foofy should be bragging about doing better than Bubbles, we need to compare their grades and calculate the difference.
Foofy earned a grade of 76, while Bubbles earned a grade of 60. The difference between their grades can be calculated by subtracting Bubbles' grade from Foofy's grade: 76 - 60 = 16.
Now, let's consider the standard deviation (SX) for each class. Foofy's standard deviation is given as SX = 10, while Bubbles' standard deviation is SX = 4.
To determine the significance of the difference between their grades, we need to calculate the z-score. The z-score measures how many standard deviations an individual's score is away from the mean (average) of the distribution.
To calculate the z-score, we use the formula: z = (x - μ) / σ, where x is the individual's score, μ is the mean, and σ is the standard deviation.
For Foofy's z-score, we have: z = (76 - μ) / 10. Since the mean is not given, we don't have enough information to calculate her z-score.
Similarly, for Bubbles' z-score, we have: z = (60 - μ) / 4. Again, since the mean is not given, we cannot calculate her z-score.
Without the means, we cannot accurately determine the significance of the grade difference between Foofy and Bubbles. Therefore, Foofy should not be bragging about doing better at this point based solely on the information provided.