Dear beloved readers, welcome to our website! We hope your visit here brings you valuable insights and meaningful inspiration. Thank you for taking the time to stop by and explore the content we've prepared for you.
------------------------------------------------ The mean IQ score of students in a particular class is 110, with a standard deviation of 5. Use the 68-95-99.7 Rule to find the percentage of students with an IQ above 120.

A. 15.85%
B. 11.15%
C. 13.5%
D. 2.5%

In this problem, we are given the mean IQ score of students in a particular class as 110, with a standard deviation of 5. We are asked to find the percentage of students with an IQ above 120 using the 68-95-99.7 Rule.

The 68-95-99.7 Rule tells us that in a normal distribution, approximately 68% of the data falls within one standard deviation of the mean, 95% falls within two standard deviations, and 99.7% falls within three standard deviations.

Since the mean IQ score is 110 and the standard deviation is 5, we can calculate the z-score for an IQ score of 120 as (120 - 110) / 5 = 2. This means that the IQ score of 120 is 2 standard deviations above the mean.

According to the 68-95-99.7 Rule, the percentage of students with an IQ above 120 is approximately 2.5%.