i took the sat and received a score of 488 on the math section. the average score was 445 with a standard deviation of 110. where does my score fit in the sample and what is the z score?

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score. Change to percentile.

On a test whose distribution is approximately normal with a mean of 50 and a standard deviation of 10, the results for three students were reported as follows:



Student Opie has a T-score of 60.

Student Paul has a z-score of -1.00.

Student Quincy has a z-score of +2.00.

Obtain the z-score and T-score for EACH student. Who did better on the test? How many standard deviation units is each score from the mean? Compare the results of the three students.

To determine where your score of 488 on the math section of the SAT fits in the sample, you can use the concept of the z-score. The z-score measures the number of standard deviations a particular data point is from the mean of a distribution.

To calculate the z-score, you can use the following formula:

z = (X - μ) / σ

Where:
- X is the value of the data point (in this case, your score of 488)
- μ is the mean of the distribution (in this case, the average score of 445)
- σ is the standard deviation of the distribution (in this case, 110)

Plugging in the values, the calculation becomes:

z = (488 - 445) / 110

z = 43 / 110

z ≈ 0.391

Now, to determine where your score fits in the sample, you need to examine the z-score in relation to the standard normal distribution (also known as the z-score table or z-table). The z-table provides the percentage or proportion of values that fall below a specific z-score.

Looking up the z-score of approximately 0.391 in the z-table, it corresponds to a proportion of approximately 0.6517. This means that around 65.17% of the sample has a score lower than 488.

Therefore, your score of 488 on the math section of the SAT is higher than approximately 65.17% of the sample.