chem 20L, how does normalization of TA grading work?

<p>I'm currently taking Chem 20L with Skibo, and I got stuck with the super-hard TA. He grades us a lot harsher than the other TAs on the lab reports, so I asked Professor Skibo what she's going to do. She says she's going to "normalize" the grades, but I'm not sure what she means by that. What does it mean? Will it help if your TA is very hard graders, and the other TAs are much easier?</p>

<p>If it's like Scerri's 20A "normalization," it will be completely arbitrary and opaque T_T</p>

<p>^ So true.</p>

<p>I didn't even knew Skibo normalized the grades. I just always thought you were screwed if you had a hard TA cause there's no curve. (I had an easy TA last quarter)</p>

<p>Yes, I'm hoping that whatever "normalizing" she does will help out the kids who have the hard TA. Still not sure what it means though :(</p>

<p>It means taking the averages of just your lab session. Then assigning a "difficulty factor" to your session compared to other sessions based on the averages of each session. The lab portion of your grade takes into consideration of the "difficulty factor". Therefore there is no grading unfairness in the relative difficulty of each session.</p>

<p>Who is this "hard TA"? Is it Kyle or Sangyoon?</p>

<p>it's Kyle lol</p>

<p>^ Hey, is he that guy with the glasses and short blonde-ish, brunette-ish (maybe reddish) hair? Lol, if he is, yeah, definitely the hard TA.</p>

<p>You should do what most of the people on CC should be doing - ask the professor directly to figure out what that process is.</p>

<p>They don't normalize for 20L.</p>

<p>I already asked the professor, and she said she will normalize the TA grading. Also, I posted my question on VOH, but I think she skipped it to answer other people's more pertinent questions pertaining to post-labs... </p>

<p>@silvercross: Kyle's hair isn't blonde, so I don't think he's the short TA lol</p>

<p>@BoelterHall: thanks!</p>

<p>so what did everyone think of the final?</p>

<p>Sucked.</p>

<p>10char</p>