Question

two students took different psychological assessments. use the given means and standard deviations to compute. 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. what is the z - score for each student? student a: z = type your answer... student b: z = type your answer... what is the proportion of students who did better than each student (above the z - score)? the proportion of students that performed better than student a is p = choose your answer... the proportion of students that performed better than student b is p = choose your answer... .0250 .0228 .9772 .0808

Explanation:

Step1: Calculate Student A's z - score

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $x$ is the 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 proportion of students better than Student A

We use the standard normal distribution table. The proportion of values to the left of $z = 1.40$ is $0.9192$. So the proportion of students better than Student A is $p_A=1 - 0.9192 = 0.0808$.

Step4: Find the proportion of students better than Student B

The proportion of values to the left of $z = 2.00$ is $0.9772$. So the proportion of students better than Student B is $p_B=1 - 0.9772=0.0228$.

Answer:

Student A: $z = 1.40$
Student B: $z = 2.00$
The proportion of students that performed better than Student A is $p = 0.0808$
The proportion of students that performed better than Student B is $p = 0.0228$