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1) A student took a chemistry exam where the exam scores were mound-shaped with a mean score of 90 and a standard deviation of 64. She also took a statistics exam where the scores were mound-shaped, the mean score was 70 and the standard deviation was 16. If the student's grades were 102 on the chemistry exam and 77 on the statistics exam, then:

a. the student did relatively better on the chemistry exam than on the statistics exam, compared to the other students in each class

b. the student's scores on both exams are comparable, when accounting for the scores of the other students in the two classes

c. it is impossible to say which of the student's exam scores indicates the better performance

d. the student did relatively better on the statistics exam than on the chemistry exam, compared to the other students in the two classes

e. the student did relatively the same on both exams

The correct answer is D but I don't know why.

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In Chemistry, she was $\frac{102-90}{64}$ standard deviation units above the mean.

In Statistics, she was $\frac{77-70}{16}$ standard deviation units above the mean. Note that $\frac{7}{16}$ is quite a bit larger than $\frac{12}{64}$, so the performance in Statistics is (comparatively) better than in Chemistry.

Measuring in standard deviation units takes above (or below) the mean takes account of the different means and different amounts of variability in the class scores in the two subjects.

"Mound-shaped" presumably means more or less normally distributed. The probability that a normal random variable with mean $90$ and standard deviation $64$ is $\ge 102$ is equal to the probability that a standard normal is $\ge \frac{112-90}{64}$. If we look at a table of the standard normal, this is approximately $0.43$. The corresponding result for Statistics is about $0.33$. Informally, that means that about $43\%$ of the students were above her in Chemistry, and only about $33\%$ were above her in Statistics.

Remark: In my experience, unmanipulated marks from Mathematics exams have a very far from normal distribution. They are often consistent with being drawn from $2$ or more populations with radically different means.

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  • $\begingroup$ . . . and probably one of those two populations consists of those who would never willingly come anywhere near a math classroom. $\endgroup$ Jun 10 '13 at 4:37
  • $\begingroup$ Thanks guys! Appreciate it. $\endgroup$
    – Andrew
    Jun 10 '13 at 8:06
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Hint: Compare each exam's corresponding $z$-scores by using the formula: $$ z = \dfrac{x-\mu}{\sigma} $$

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