Question

As in the previous question, scores on a certain test follow the normal curve with an average of 1350 and a standard deviation of 120. in order to qualify for a certain job, a candidate needs to score in the top 2.5%. what score does she need?

Answer 1

To qualify for the job by scoring in the top 2.5% of a test with a normal distribution, a mean of 1350, and a standard deviation of 120, a candidate needs a score of approximately 1585.

Step-by-step explanation:

To find the score a candidate needs to score in the top 2.5% on a test with a normal distribution, first, we need to determine the z-score that corresponds to the top 2.5% of the distribution. Consulting a standard normal distribution table, we find that the z-score that leaves 2.5% in the upper tail is approximately 1.96. Once we have the z-score, we can use the formula for a score in a normal distribution:

Score = Mean + (Z * Standard Deviation)

Given that the mean score is 1350 and the standard deviation is 120, we can substitute these values into the formula along with our z-score.

Score = 1350 + (1.96 * 120)
Score = 1350 + 235.2
Score = 1585.2

To qualify for the job, a candidate would therefore need to score approximately 1585 or higher on the test.