Question

comparing student test scores using z - scores
as part of a psychological assessment course, two students—student a and student b—completed different sta performances relative to the average scores on each test by calculating and interpreting z - scores and associate

• student a scored 82 on a test with a mean of 75 and a standard deviation of 5.
• student b scored 140 on a different test with a mean of 100 and a standard deviation of 20.

although they took different tests, both are normally distributed. your task is to:

1. calculate each students z - score
2. use the z - table to find the probability of scoring higher than each student
3. interpret the results in apa style using probability - based language

compute two z - scores
instructions:
two students took different psychological assessments. use the given means and standard deviations to compute ea
please round your answers to two decimal places.
student a scored 82 on a test with a mean of 75 and a standard deviation of 5.
student b scored 140 on a test with a mean of 100 and a standard deviation of 20.

Explanation:

Step1: Calculate Student A's z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the raw score, $\mu$ is the mean, and $\sigma$ is the standard deviation. For Student A, $x = 82$, $\mu=75$, and $\sigma = 5$. So $z_A=\frac{82 - 75}{5}=\frac{7}{5}=1.40$

Step2: Calculate Student B's z - score

For Student B, $x = 140$, $\mu = 100$, and $\sigma=20$. Using the z - score formula $z=\frac{x-\mu}{\sigma}$, we have $z_B=\frac{140 - 100}{20}=\frac{40}{20}=2.00$

Step3: Find the probability of scoring higher than Student A

Looking up the z - score of $z_A = 1.40$ in the z - table, the cumulative probability $P(Z\leq1.40)$ is approximately $0.9192$. The probability of scoring higher than Student A is $P(Z > 1.40)=1 - 0.9192 = 0.0808$

Step4: Find the probability of scoring higher than Student B

Looking up the z - score of $z_B=2.00$ in the z - table, the cumulative probability $P(Z\leq2.00)$ is approximately $0.9772$. The probability of scoring higher than Student B is $P(Z > 2.00)=1 - 0.9772=0.0228$

Step5: Interpret the results in APA style

Student A's z - score of $z = 1.40$ indicates that their score was 1.40 standard deviations above the mean. The probability of a randomly - selected student scoring higher than Student A is $p = 0.0808$. Student B's z - score of $z = 2.00$ indicates that their score was 2.00 standard deviations above the mean. The probability of a randomly - selected student scoring higher than Student B is $p = 0.0228$. This suggests that Student B performed relatively better compared to Student A within their respective test populations, as a smaller proportion of students are expected to score higher than Student B.

Answer:

1. Student A's z - score: $1.40$, Student B's z - score: $2.00$
2. Probability of scoring higher than Student A: $0.08$; Probability of scoring higher than Student B: $0.02$
3. Student A's score was 1.40 standard deviations above the mean, with a probability of 0.08 of a student scoring higher. Student B's score was 2.00 standard deviations above the mean, with a probability of 0.02 of a student scoring higher. Student B performed relatively better within their test population.