Question

The scores on the entrance exam at a well-known, exclusive law school are normally distributed with a mean score of 140 and a standard deviation equal to 43. At what value should the lowest passing score be set if the school wishes only 2.5 percent of those taking the test to pass? (Round your answer to nearest whole number.

Answer 1

To determine the lowest passing score, we find the equivalent z-score for the 97.5th percentile and apply it to the entrance exam's normal distribution with a mean of 140 and a standard deviation of 43. This value is approximately 224 when rounded to the nearest whole number.

Step-by-step explanation:

To find the value for the lowest passing score on the entrance exam at the exclusive law school, where only 2.5 percent of those taking the test pass, we need to look at the upper tail of the normal distribution.

Given that the mean score is 140 and the standard deviation is 43, we are looking for the z-score that corresponds to the 97.5th percentile (100% - 2.5%).

Using a standard normal distribution table or a calculator with the inverse cumulative distribution function for the normal distribution, we find that the z-score that corresponds to the 97.5th percentile is approximately 1.96.

Using the formula for converting a z-score to the original score (X), which is: X = mean + z * standard deviation, we calculate the passing score as follows:

X = 140 + 1.96 * 430

≈ 224

So, the lowest passing score should be set at approximately 224, rounding to the nearest whole number.