I have a problem on standard deviation, and I will need some help with it.
Problem:
Each professor in an English class give an A to students scoring above 90 and a F to those scoring below 60.
Professor 1 has a normal distribution with a mean of 74 and standard deviation 7
Professor 2 has a normal distribution with a mean of 78 and standard deviation 18.
Which professor gives the higher proportion of A's?
Thank you for your help!)
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the two Z scores.
To determine which professor gives a higher proportion of A's, we need to calculate the area under the curve of each professor's distribution that corresponds to scores above 90.
To do this, we can use the z-score formula:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value we want to calculate the proportion above (90 in this case)
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
First, let's calculate the z-score for Professor 1:
z1 = (90 - 74) / 7
z1 ≈ 2.29
Using a z-score table or statistical software, we can find the proportion of the distribution above this z-score. For Professor 1, this represents the proportion of students who scored above 90.
Now, let's calculate the z-score for Professor 2:
z2 = (90 - 78) / 18
z2 ≈ 0.67
Again, using a z-score table or statistical software, we can find the proportion of the distribution above this z-score. For Professor 2, this represents the proportion of students who scored above 90.
Finally, compare the proportions obtained for both professors. The professor with the higher proportion of A's is the one with the greater proportion above 90.
Note: The proportion above a z-score can be found using a z-score table or by using statistical software like R, Python's scipy.stats library, or Excel's NORM.S.DIST function.