I am trying to find out the cut off score for selecting 10% of applicants with a mean of 500 and a standard deviation of 100 assuming that the standadized test is normally distributed.
To find the cut off score for selecting 10% of applicants, you need to determine the z-score that corresponds to the 10th percentile of the standard normal distribution.
1. Start by finding the z-score using the cumulative distribution function (CDF) of the standard normal distribution. The z-score represents the number of standard deviations away from the mean.
2. Use a z-table or a statistical calculator to find the z-score that corresponds to the 10th percentile. In this case, since we want to select the top 10% of applicants, we are looking for the z-score that leaves 10% of the distribution to the left.
3. Once you have the z-score, you can calculate the raw score using the formula:
raw score = (z-score * standard deviation) + mean
In this specific scenario, you mentioned that the mean is 500 and the standard deviation is 100. Therefore, we need to find the z-score for the 10th percentile.
Using a z-table or a statistical calculator, you can find that the z-score corresponding to the 10th percentile is approximately -1.28.
To find the cut off score:
cut off score = (-1.28 * 100) + 500 = 372
Therefore, the cut off score for selecting 10% of applicants is 372.