SOLUTION: A student scores 60 on a Mathematics test that has a mean of 54 and a standard deviation of 3, and she scores 80 on a history test with a mean of 75 and a standard deviation of 2.

Question 1171133: A student scores 60 on a Mathematics test that has a mean of 54 and a standard deviation of 3, and she scores 80 on a history test with a mean of 75 and a standard deviation of 2. On which test did she perform better?

Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
use the z-score formula to determine the answer to this.
z-score formula is:
z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation

math test has mean of 54 with standard deviation of 3.
z = (60 - 54) / 3 = 6/3 = 2.
area to the left of that z-score = .977249938.
round to 3 decimal places to get .977
multiply by 100 to get z-score of 2 is greater than 97.7% of all possible z-scores.

history test has mean of 75 with standard deviation of 2.
z = (80 - 75) / 2 = 5/2 = 2.5
area to the left of that z-score = .9937903201.
round to 3 decimal places to get .994
multiply by 100 to get z-score of 2.5 is greater than 99.4% of all possible z-scores.

she performed better on the history exam.

i used the t-84 plus to find the probabilities.
if you used the z-score table, you would do the following.
look up the z-score (rounded to 2 decimal places) in the table to find the area under the normal distribution curve to the left of that z-score.

with a z-score of 2.00, the area to the left of the z-score is shown as .97725 = .977 rounded to 3 decimal places * 100 = 97.7%.
with a z-score of 2.50, the area to the left of the z-score is shown as .99379 = .994 rounded to 3 decimal places * 100 = 99.4%.

the table i used can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf

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