Question 1169595: Hello, I need help solving the following problem (please show the steps):
In 2005, 1,475,623 students heading to college took the SAT. The distribution of scores in the math section of the SAT follows a normal distribution with mean
μ = 520 and standard deviation σ = 115.
For 2012, the SAT math test had a mean of 514 and a standard deviation of 117. The ACT math test is an alternative to the SAT and is approximately normally distributed with a mean of 21 and a standard deviation of 5.3. If one person took the SAT math test and scored 700 and a second person took the ACT math test and scored 34, who did better with respect to the test they took? (answers must be rounded to two decimal places)
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! in 2005, the sat test had a mean of 520 and a standard deviation of 115.
in 2012, the sat test had a mean of 514 and a standard deviation of 117.
the act test had a mean of 21 and a standard deviation of 5.3.
sat score was 700.
act score was 34.
z-score = (x - m) / s
x is the raw score
m is the mean
s is the standard deviation
for the sat in 2009, z = (700 - 520 / 115 = 1.57
for the sat in 2012, z = (700 - 514) / 117 = 1.59
for the act, z = (34 - 21) / 5.3 = 2.45
the person who took the act test scored significantly better compared to the other people who took the act test.
with a z-score of 1.57, the score was better than 94% of the other people who took the test.
with a z-score of 1.59, the score was better than 94% of the other people who took the test.
with a z-score of 2.45, the score was better than 99% of the people who took the test.
the person who took the act test did better in comparison to the other people who took the act test than the person who took the sat test did in comparison to the other people who took the sat test.
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