Here it is word for word...

Professor Brown's morning class has 66 students and his evening class has 45 students. On the first exam, everyone in the morning class got a 68 while the evening class averaged a 72 and had a (Sumxi squared) =240,610. Suppose Professor Brown combined these two classes into one.

1.) What would be the combined class mean? (69.6216) my answer

2.) What would be the combined class variance? ????

3.) What is the median? What is the shape of the distribution? (Median=68 and is skewed right) my answer

Any help..

 

Everyone takes a peek, but no one helps....

 

I haven't done statisitcs in a while, but I think this is mostly right. . .

1. Your answer to number one is correct = (66*68)+(45*72)/(45+66)

2. The variance = the s^2, which is I think is ((68 - 69.6216)^2 + (72 - 69.6216)^2)/111 = 0.714

3. Without crunching the numbers, I think your answer for 3 is correct. . .

 

Thanks a bunch.. I'm still working on #2.. Thanks again

 

I'm having a hard time understanding why #2 is .714. Shouldn't the denominator be n-1 , which would be 110?? And If I punch in your calculation I keep getting .0747

 

OMG!! Stop, stop! I'm getting a headache already.
What you posted is like the most disgusting thing I've read on here!

 

i didnt logon to think!

 

I feel the same way.. Too bad it's due tomorrow..

 

Variance is E((X - mu)^2). Using some algebra:

E((X - mu)^2) =

E(X^2 - 2*X*mu + mu^2) =

E(X^2) - E(2*X*mu) + E(mu^2) =

E(X^2) - 2*mu*E(X) + mu^2 =

E(X^2) - 2*mu*mu + mu^2 =

E(X^2) - mu^2.

E(X^2) = sum(X^2) / n.

You are given sum(X^2) = 240,610 for the evening class.

Sum(x^2) = 305,184 for the morning class. (Why?)

So, E(X^2) = (240,610 + 305,184) / 111 = 4917.06

mu^2 = 4847.17

So, variance = 4917.06 - 4847.17 = 69.90

 

Perfect.. Thank you soooo much.. Now I can have a peaceful dinner tonight.. I owe you one..