Section A has 30 students with mean 54 and section B has 45 students; the two together average 60. Later a score of 82 in section B is found to have been entered as 28. What is the corrected mean of section B?

Question

Accepted Answer

65.2. Total =75×60=4500; A contributes 1620, so B's recorded total is 2880 (mean 64). Correcting: 2880-28+82=2934, mean =2934/45=65.2. (64 is the uncorrected mean.)

Section A has 30 students with mean 54 and section B has 45 students; the two together average 60. Later a score of 82 in section B is found to have been entered as 28. What is the corrected mean of section B?

Step-by-Step Solution

Master this topic in 30 minutes a day.

More Statistics practice