Final answer:
By calculating the z-score for a score of 350 on an exam with a mean of 500 and a standard deviation of 75, we get a z-score of -2, which corresponds to the 2.5th percentile. However, since the given options don't include this percentage, option (a) 16%, being the closest and smallest, is likely the correct answer for what percent of students scored below 350.
Step-by-step explanation:
To find what percent of the students scored below 350 on a college entrance exam, where the mean score is 500 and the standard deviation is 75, we use the properties of the normal distribution. First, we calculate the z-score for a score of 350:
Z = (X - µ) / σ
Where:
X = target score (350)
µ = mean score (500)
σ = standard deviation (75)
Substituting the values:
Z = (350 - 500) / 75
Z = -150 / 75
Z = -2
A z-score of -2 corresponds to the 2.5th percentile in a standard normal distribution, which implies that approximately 2.5% of scores fall below this z-score. However, the options provided in the question (16%, 32%, 42%, 68%) don't include this percentage, which suggests that either the z-score calculation or the interpretation of percentiles in the normal distribution needs to be considered differently. For typical educational assessments and assuming a bell curve distribution, usually about 16% fall below one standard deviation (-1 z-score). However, since -2 is more than one standard deviation away from the mean, a smaller percentage of students would score this low. Since none of the options are less than 16% and 16% is the smallest percentage option provided, the most likely correct answer would be (a) 16%.