Hello, thank you for taking the time to post your question!
No, it’s inside of one standard deviation of the mean value so it would not be considered unusual
Z = (82.3 – 77.1) / 12.6 = 5.2 / 12.6 = 0.41 , meaning that it is well within the typical range for this group
I hope that helps you get moving in a better direction on this type of question! Feel free to reach out if you have any additional questions beyond that :)