Question

comparing test scores for the statistics test scores in exercise 37, the mean is 63 and the standard deviation is 7.0, and for the biology test scores in exercise 38, the mean is 23 and the standard deviation is 3.9. in exercises 39 - 42, you are given the test scores of a student who took both tests. (a) transform each test score to a z - score.

Explanation:

Step1: Recall z - score formula

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the raw score, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Assume a student's statistics score is $x_{1}$

Let the student's statistics score be $x_{1}$. The mean of statistics scores $\mu_{1}=63$ and standard deviation $\sigma_{1} = 7.0$. Then the z - score for the statistics test, $z_{1}=\frac{x_{1}-63}{7.0}$.

Step3: Assume a student's biology score is $x_{2}$

Let the student's biology score be $x_{2}$. The mean of biology scores $\mu_{2}=23$ and standard deviation $\sigma_{2}=3.9$. Then the z - score for the biology test, $z_{2}=\frac{x_{2}-23}{3.9}$.

Answer:

The z - score for the statistics test is $z_{1}=\frac{x_{1}-63}{7.0}$ and the z - score for the biology test is $z_{2}=\frac{x_{2}-23}{3.9}$, where $x_{1}$ is the student's statistics score and $x_{2}$ is the student's biology score.