I am faced with the above question, I'm aware that a linear combination of a normally distributed variable maintains its normality, however in this question it does not seem to be facing a linear combination. Nor did it first resemble a chi-squared to me.
That said my original (and probably very incorrect) working went something like:
$$Y=\sum_{i=1}^{N}\sigma^{-2}(X_i-\mu)^2$$ $$=\sigma^{-2}\sum_{i=1}^{N}(X_i-\mu)^2$$
As $$\sigma^2=\frac{1}{N-1}\sum_{i=1}^{N}(X_i-\mu)^2$$ then $$\sum_{i=1}^{N}(X_i-\mu)^2=(N-1)\sigma^2$$
Therefore $$ Y=\sigma^{-2}(N-1)\sigma^2 $$
$$Y=N-1$$ which is obviously just a constant, and I imagine completely wrong. Intuitively Y can't be negative, and therefore can't be normally distributed around 0, unless it's variance was 0 (though I'm not sure that would still be considered a normal distribution?).
And therefore my best guess would be it's (c) - which I can sort of understand, $\sum_{i=1}^{N}\frac{(X_i-\mu)^2}{\sigma^2}$ effectively acts as a way to standardise the distribution, and it's then squared and summed, giving a Chi-squared with N degrees of freedom.
Any guidance on this would be greatly appreciated.
self-studytag and read its tag-wiki. In approaching this, first consider what happens for $N=1$ $\endgroup$ – Glen_b Sep 21 '15 at 16:00