SOLUTION: The marks in English in a recent test had a mean of 76% with a standard deviation of 7%. The Science results had a mean of 84% with a standard deviation of 5%. Jasmine obtained sco

Question 1191685: The marks in English in a recent test had a mean of 76% with a standard deviation of 7%. The Science results had a mean of 84% with a standard deviation of 5%. Jasmine obtained scores of 86% in English and 89% in Science.
a) Which is the better result? Explain your answer.
b) If there are 30 people in the class, estimate Jasmine's position in Science.
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the marks in english have a mean of 76% with a standard deviation of 7%.
the marks in science have a mean of 84% with a standard deviation of 5%.

jasmine obtained 86% in english and 89% in science.

use the z-score to compare these.

the z-score formula says z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation, in this case.

for english, her z-score formula becomes z = (86 - 76) / 7 = 1.43, rounded to 2 decimal places.
for science, her z-score formula becomes z = (89 - 84) / 5 = 1, no additional rounding required.

a z-score of 1.43 has a greater area under the normal distribution curve to the left of it.
this indicates she scored higher, relative to her peers, in the english course.
while she had a higher score in science, the rest of the class did better too, making her score less higher, relative to her peers.

fyi, a z-score of 1.43 has an area under the normal distribution curve equal to
.92364 to the left of it and a z-score of 1 has an area under the normal distribution curve equal to .84134.

an area of .92364 to the left of the z-score means she was better than approximately 92% of her peers in english.
an area of .84134 to the left of the z-score means she was better than approximately 84% of her peers in science.

let me know if you have any questions.
theo

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