I am trying this exam question and I have no idea where to start for the final part (part iv), because I am not sure how you get sigma x^2 or the mean, using the values, because I don't know why the 200 is being used in the equations.
If you could provide any advice on solving this problem, I'll be grateful (please explain your answer as simply as possible and if there are any issues with the question, please let me know).
On
\begin{eqnarray*} \sum_{i=1}^{n}\left( x_{i}-200\right) =245 \end{eqnarray*} \begin{eqnarray*} \sum_{i=1}^{n}\left( x_{i}-200\right) ^{2}=9849 \end{eqnarray*} \begin{eqnarray*} n=49 \end{eqnarray*} \begin{eqnarray*} \bar{x} &=&\frac{\sum_{i=1}^{n}x_{i}}{n}=\frac{245+200\times 49}{49} \\ &=&205\not=200 \end{eqnarray*} \begin{eqnarray*} \left( x_{i}-\bar{x}\right) ^{2} &=&\left( x_{i}-200+200-\bar{x}\right) ^{2} \\ &=&\left( x_{i}-200\right) ^{2}+2\left( x_{i}-200\right) \left( 200-\bar{x} \right) +\left( 200-\bar{x}\right) ^{2} \\ &=&\left( x_{i}-200\right) ^{2}-10\left( x_{i}-200\right) +25 \\ &=&\left( x_{i}-200\right) ^{2}-10x_{i}+2025 \end{eqnarray*} \begin{eqnarray*} \sum_{i=1}^{n}\left( x_{i}-\bar{x}\right) ^{2} &=&\sum_{i=1}^{n}\left( x_{i}-200\right) ^{2}-10\sum_{i=1}^{n}x_{i}+2025\sum_{i=1}^{n}1 \\ &=&\sum_{i=1}^{n}\left( x_{i}-200\right) ^{2}-10n\bar{x}+2025n \\ &=&9849-10\times 49\times 205+2025\times 49 \\ &=&8624 \end{eqnarray*} \begin{eqnarray*} s^{2} &=&\frac{1}{n-1}\sum_{i=1}^{n}\left( x_{i}-\bar{x}\right) ^{2} \\ &=&\frac{8624}{48}=179.67 \end{eqnarray*} \begin{eqnarray*} s &=&\sqrt{179.67} \\ &=&13.404 \end{eqnarray*}
The way I would do this is to define $Y=X-200$. Then you know $\sum Y$ and $\sum Y^2$, so it's easy to calculate the mean and standard deviation of $Y$; and the mean of $X$ is the mean of $Y$ plus $200$, while the standard deviation of $X$ is the same as the standard deviation of $Y$.