given that $K_{1}$, $K_{2}$, ... , $K_{n}$ is a random variables from a normal distribution with $μ$ = 10, $σ^2$ = 2.5
what is the distribution and parameter for $X$ = $\sum_{i=1}^{10} \left(\frac{K_{i} - 10}{\sqrt{2.5}}\right)$ ?
is $x̄$~$N$(10,0.25) the right answer ? or do i need to use the chi-square distribution ?
Hope this helps:
Knowing that: $K_{j} \sim N(10,2.5), \forall j \in {1,2,..,n}$
We want to determinate the distribution of \begin{equation*} X = \sum_{i=1}^{10} (\frac{K_{i}-10}{\sqrt{2.5}}) \end{equation*} Please note that for each $\underline{i}$, $$\frac{K_{i}-10}{\sqrt{2.5}}\sim N(0,1)$$ So, rewriting X, we have: \begin{equation*} X = Z_{1} + Z_{2} + ... + Z_{10}, \text{ where } Z_{i} \sim N(0,1), i \in {1,2,...,10} \end{equation*} Now, if the variables are independent, you get a sum of Normal Distributed variables which is a Normal Distributed Variable.
You could use the Chi-Square distribution if the standard Z variables were squared up, this is, if: \begin{align*} X = \sum_{i=1}^{10} (\frac{K_{i}-10}{\sqrt{2.5}})^{2} \end{align*}