AS Maths Standard Deviation Question

Stuck on the same question! Did you figure it out?

Have you tried writing out the variance formula? Please post what you've tried.

So I wrote that the variance is sigma x^2 -121^2 = 226 so sigma x^2 = (226+121^2) x 9 so sigma x^2 = 122803. I also just wrote down that the total IQ is 121 x 9 so sigma x is 1089. These are all the calculations I managed to do with the first set of data.

I then wrote down that new variance is sigma x^2 (different from the previous one) divided by 10 (as there are now 10 valuues) - (sigma x divided by 10)^2 is equal to 239.4.

I don't know where to go from there

Call the new student's IQ $y$.

Then the new sum of the squares is going to be $\Sigma x^2 + y^2$. Does this make sense?

See if you can continue from here.

That does make sense! I have tried but I don't know whether I am approaching the question incorrectly?

So I rewrote out the variance as (sigma x^2 + y^2)/10 - (1089 + y)/10 is equal to 239.4 so (sigma x^2 + y^2) - (1089 + y) is equal to 2394. As you can see above, from my previous calculations I got the initial sum of squares as 133803 so I substituted this in. So 133803 + y^2 - 1089 - y = 2394 so y^2 - y = -130320 but as soon as I got a negative I was pretty sure I am doing it incorrectly?

I just don't know how to go about answering this question. I've played around with it a bit and seem to be getting no where. Would appreciate it if someone could help me out.

The question:
The mean of the IQ of a class of 9 students is 121 and the variance is 226. Another student joins the class and the variance changes to 239.4 .
What are the possible values of the IQ of the new student?

At first glance I can't see anything wrong but that quadratic doesn't have solutions. I recommend trying it again and see if you can correct your mistake.

I can try the question myself a little bit later but I'm trying to do a few things at once right now - sorry for the delayed replies

That does make sense! I have tried but I don't know whether I am approaching the question incorrectly?

So I rewrote out the variance as (sigma x^2 + y^2)/10 - (1089 + y)/10 is equal to 239.4 so (sigma x^2 + y^2) - (1089 + y) is equal to 2394. As you can see above, from my previous calculations I got the initial sum of squares as 133803 so I substituted this in. So 133803 + y^2 - 1089 - y = 2394 so y^2 - y = -130320 but as soon as I got a negative I was pretty sure I am doing it incorrectly?

Approach is good, but:

$Var(X)=\mathbb{E}(X^2)-(\mathbb{E}(X))^2$

You seem to have done:

$Var(X)=\mathbb{E}(X^2)-\mathbb{E}(X)$