The graduate selection committee wants to select the top 10% of applicants. On a standardized test with a mean of 500 and a standard deviation of 100, what would be the cutoff score for selecting the top 10% of applicants, assuming that the standardized test is normally distributed?
To find the cutoff score for selecting the top 10% of applicants on a standardized test, we need to use the standard normal distribution table or calculator.
Here are the steps to find the cutoff score:
1. Start by converting the desired percentage (10%) into a z-score. We need to find the z-score that corresponds to the area to the left of the cutoff score.
2. The z-score represents the number of standard deviations from the mean. We need to find the z-score that corresponds to the area to the left of the desired percentage (10%).
3. Open a standard normal distribution table or use a calculator with a standard normal distribution function to find the z-score corresponding to the desired percentage. The area to the left of the z-score is equal to the desired percentage (10%).
4. Once you find the z-score, use the formula z = (x - μ) / σ to convert the z-score into the corresponding raw score (x). Plug in the values for the z-score, mean (μ), and standard deviation (σ).
In this case, the mean is 500 and the standard deviation is 100. We want to find the cutoff score for the top 10% of applicants.
Let's go through the steps:
1. To find the z-score corresponding to the top 10% (or 0.10), we need to find the inverse of the cumulative distribution function (CDF). In other words, we need to find the critical z-value for which the area to the left is 0.10.
2. Using a standard normal distribution table or calculator, we find that the critical z-value is approximately 1.28.
3. Now, we can use the formula z = (x - μ) / σ to find the raw score (x).
Rearranging the formula, we have x = z * σ + μ.
Plugging in the values, we have x = 1.28 * 100 + 500 = 628.
Therefore, the cutoff score for selecting the top 10% of applicants is 628 on the standardized test.