Final answer:
Approximately 15.87% of the students passed the test, calculated by finding the z-score for the passing mark of 50%, which is 1 standard deviation above the mean, and using the standard normal distribution to determine the proportion of students who scored higher.
Step-by-step explanation:
To find out what percentage of students pass the test, given the mean score is 42% with a standard deviation of 8% on a normally distributed curve, and the passing mark is 50%, we use the concept of the normal distribution. We need to calculate the z-score for a test score of 50% to determine how many standard deviations this score is from the mean.
Steps to Calculate the Z-Score
Subtract the mean from the test score: 50% - 42% = 8%.
Divide the result by the standard deviation: 8% / 8% = 1.
This means that a test score of 50% is 1 standard deviation above the mean. To find the percentage of students who scored above this z-score, we need to look at standard normal distribution tables or use a calculator that provides the cumulative area under the normal curve to the left of a given z-score.
From the standard normal distribution, we find that the cumulative area to the left of a z-score of +1 is approximately 0.8413. This represents the proportion of students who scored below the passing mark. To find the percentage of students who passed (scoring 50% or above), we subtract this value from 1.
1 - 0.8413 = 0.1587, which means that approximately 15.87% of the students passed the test.
To answer the question, approximately 15.87% of students passed the test, based on the provided information and assuming the data is normally distributed.