Question

Out of the students who attempted the midterm for Ms.Wilson's IM3 class last October, the mean score was 80% with a standard deviation of 9%. It was approximately normally distributed. About what percent of students scored between a 71% and an 89%? About what percent scored a 75% or more?

Answer 1

To find the percentage of students who scored between 71% and 89%, calculate the z-scores and use the standard normal distribution table. The percentage is approximately 68.27%. To find the percentage of students who scored 75% or more, calculate the z-score and find the area to the right in the standard normal distribution table. The percentage is approximately 70.88%.

Step-by-step explanation:

To find the percentage of students who scored between 71% and 89%, we need to calculate the z-scores for these values and then use the standard normal distribution table. The z-score for 71% is calculated as (71 - 80) / 9 = -1, and the z-score for 89% is (89 - 80) / 9 = 1. Using the standard normal distribution table, we can find that the area between -1 and 1 is approximately 0.6827. Multiply this by 100 to get the percentage, so about 68.27% of students scored between 71% and 89%.

To find the percentage of students who scored 75% or more, we need to calculate the z-score for 75% using the formula (75 - 80) / 9 = -0.5556. Looking this up in the standard normal distribution table, we find that the area to the right of -0.5556 is approximately 0.7088. Multiply this by 100 to get the percentage, so about 70.88% of students scored 75% or more.