Question

question 3
on a recent biology midterm, the class mean was 73 with a standard deviation of 2.2. calculate the z - score (to 4 decimal places) for a person who received score of 77.
z - score for biology midterm.
the same person also took a midterm in their marketing course and received a score of 81. the class mean was 78 with a standard deviation of 5.8. calculate the z - score (to 4 decimal places).
z - score for marketing midterm.
which test did the person perform better on compared to the rest of the class? select an answer
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Explanation:

Step1: Recall z - score formula

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

Step2: Calculate z - score for biology mid - term

For the biology mid - term, $x = 77$, $\mu=73$, and $\sigma = 2.2$. Substitute these values into the formula: $z_{biology}=\frac{77 - 73}{2.2}=\frac{4}{2.2}\approx1.8182$.

Step3: Calculate z - score for marketing mid - term

For the marketing mid - term, $x = 81$, $\mu = 78$, and $\sigma=5.8$. Substitute into the formula: $z_{marketing}=\frac{81 - 78}{5.8}=\frac{3}{5.8}\approx0.5172$.

Step4: Compare z - scores

Since $1.8182>0.5172$, the person performed better on the biology mid - term compared to the rest of the class.

Answer:

z - score for Biology Midterm: 1.8182
z - score for Marketing Midterm: 0.5172
The person performed better on the biology mid - term.