# Statistics - Lost with this question

I'm having trouble doing this question because I don't know where to begin. Could someone walk me through this slowly so that I understand the thought process and how to approach questions like this?

Let $X$ be the weight (in grams) of a nail of the type that is used for making decks. Assume that the distribution of X is Normal $X\sim N(\mu=8.78,\, \sigma^2=0.16)$. Let $\bar{X}$ be the mean of a random sample of the weights of $n = 9$ nails.

(a) Sketch, on the same set of axes, the graphs of the probability density functions of X and of $\bar{X}$. (b) Let $S^2$ be the sample variance of the nine weights. Find constants $a$ and $b$ so that $P(a < S^2 < b) = 0.90$. (Please look for $a$ and $b$ such that $P(S^2 < a) = 0.05$ and $P(S^2 > b) = 0.05$.)

Thank you.

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(a)The distribution of $\bar{X}$ for $n=9$ in Normal with mean $\mu_n=8.78$ (the same as $X$) and variance $\sigma_n^2=\frac{0.16}{n}$. That follows from the additive properties of the normal distribution. Now using that the probability density function of the normal distribution is $$f(x)=\frac{1}{\sqrt{2\pi\sigma^2}}\cdot e^{-\frac{(x-\mu)^2}{2\sigma^2}}$$ you can draw the graphs (by using an appropriate software for example) of $X$ and $\bar{X}$ by substituting $\mu, \sigma$ for $X$ and $\mu_n, \sigma_n$ (as they are stated above) for $\bar{X}$. If you do not use a software, the draw two normal curves (see Wikipedia for the exact shape), with mean $8.78$ and the one with the bigger variance (i.e. shorter and flater) will be the one for $X$.
For b) you need that the sample variance $S^2$ has the chi-squared distribution with $n-1=8$ degrees of freedom. You need to determine the points $\chi^2_{8;0.05}$ and $\chi^2_{8;0.95}$ from the chi squared distribution which can be found in the tables at the back of your book (I guess). They are equal to: $$a=\chi^2_{8;0.05}=2.733$$ and $$b=\chi^2_{8;0.95}=15.507$$