Assuming SAT score data are normally distributed, if only the top 25% of students based on SAT scores qualify to be interviewed for a particular school what would be the minimum score required to have to be interviewed if the population mean is a 1509 and the standard deviation is 113
from your standard Z table, the top 25% is above µ+0.674σ.
So, that would be 1509+.674*113 = 1585
To find the minimum score required to be interviewed, we need to determine the cut-off score that separates the top 25% of students from the rest of the population.
Step 1: Find the z-score corresponding to the desired percentile.
The top 25% corresponds to the 75th percentile. We can find the z-score using a standard normal distribution table or a calculator. In this case, we want to find the z-score for the 75th percentile, which is denoted as z(0.75).
Step 2: Convert the z-score to the corresponding SAT score.
Since we know the population mean (1509) and the standard deviation (113), we can use the formula to convert the z-score to the original SAT score:
SAT score = (z-score * standard deviation) + mean
Let's calculate the minimum score required:
Step 1: Find the z-score corresponding to the 75th percentile.
You can use a standard normal distribution table or a calculator. The z-score for the 75th percentile is approximately 0.6745.
Step 2: Convert the z-score to the corresponding SAT score.
SAT score = (0.6745 * 113) + 1509
Calculating the above expression:
SAT score ≈ 76.3235 + 1509
SAT score ≈ 1585.3235
Therefore, the minimum SAT score required to be interviewed for the particular school is approximately 1585.