I have deduced this from the question:
We have $9$ independent and identically distributed (i.i.d.) random variables $X_1, X_2, ..., X_9$, each following a normal distribution with mean $μ$ and variance $1$ ($N(μ, 1)$). We want to find the joint probability that:
$(\bar{X} - μ)^2$
$\sum_{i=1}^9(X_i - \bar{X})^2$
where $\bar X$ is the sample mean.
I know that $\sum_{i=1}^9(X_i - \bar{X})^2$ will follow chi-squared distribution with $8$ degrees of freedom ($χ^2(8)$). But I don't know what $(\bar{X} - μ)^2$ will follow. Will the two distributions be independent?
$\def\ed{\stackrel{\text{def}}{=}}$ You can simplify matters a little by setting $\ Y_i\ed X_i-\mu\ .$ Then $\ Y_i\ $ are standard normal variates with \begin{align} \bar{Y}&=\bar{X}-\mu\\ \text{Var}(\bar{Y})&=\text{Var}(\bar{X})=\mathbb{E}\big(\bar{Y}^2\big)=\frac{1}{9}\\ \sum_{i=1}^9\big(X_i-\bar{X}\big)^2&=\sum_{i=1}^9\big(X_i-\mu-(\bar{X}-\mu)\big)^2\\ &=\sum_{i=1}^9\big(Y_i-\bar{Y}\big)^2\ , \end{align} and the probability you want is $$ \mathbb{P}\left(\bar{Y}^2\ge\frac{1}{9}, \sum_{i=1}^9\big(Y_i-\bar{Y}\big)^2\ge4\right)\ . $$ Now since $$ \pmatrix{Y_1-\bar{Y}\\Y_2-\bar{Y}\\\vdots\\Y_9-\bar{Y}\\\bar{Y}}=\pmatrix{I_{9\times9}-\frac{1}{9}\mathbf{1}_{9\times1}\mathbf{1}_{9\times1}^T\\\frac{1}{9}\mathbf{1}_{9\times1}^T}\pmatrix{Y_1\\Y_2\\\vdots\\Y_9}\ , $$ is a linear function of the vector $\ \pmatrix{Y_1&Y_2&\dots&Y_9}\ $ of independent standard normal variates, then the random variates $\ Y_1-\bar{Y},$$\,Y_2-\bar{Y},$$\,\dots,Y_9-\bar{Y},$$\,\bar{Y}\ $ follow a (degenerate) multivariate normal distribution with mean $\ \pmatrix{0&0&\dots&0}\ $ and covariance matrix $$ \pmatrix{I_{9\times9}-\frac{1}{n}\mathbf{1}_{9\times1}\mathbf{1}_{9\times1}^T&\mathbf{0}_{9\times1}\\\mathbf{0}_{9\times1}^T&\frac{1}{9}}\ . $$ Since $\ \text{Cov}\big(Y_i-\bar{Y},\bar{Y}\big)=0\ ,$ for every $\ i\ ,$ it follows that $\ \bar{Y}\ $ is indeed independent of $\ Y_1-\bar{Y},$$\,Y_2-\bar{Y},$$\,\dots,Y_9-\bar{Y}\ $ and hence also of $\ \sum_\limits{i=1}^9\big(Y_i-\bar{Y}\big)^2\ ,$ and therefore $$ \mathbb{P}\left(\bar{Y}^2\ge\frac{1}{9}, \sum_{i=1}^9\big(Y_i-\bar{Y}\big)^2\ge4\right)=\mathbb{P}\left(\bar{Y}^2\ge\frac{1}{9}\right)\mathbb{P}\left(\sum_{i=1}^9\big(Y_i-\bar{Y}\big)^2\ge4\right)\ . $$ Since $\ \bar{Y}\ $ is normally distributed with mean $\ 0\ $ and variance $\ \frac{1}{9}\ ,$ it follows that $\ 9\bar{Y}^2\ $ follows a chi-squared distribution with one degree of freedom.