The table shows the probability distribution of a random variable X, which can only take the values $1,2,3,4$ and 5.

\begin{center}
\begin{tabular}{ c | c | c | c | c | c }
k & 1 & 2 & 3 & 4 & 5 \\
\hline
$\mathrm { P } ( \mathrm { X } = \mathrm { k } )$ & $\mathrm { p } _ { 1 }$ & $\mathrm { p } _ { 2 }$ & $\mathrm { p } _ { 3 }$ & 0,2 & 0,15 \\
\hline
\end{tabular}
\end{center}

The probabilities $P ( X = 4 )$ and $P ( X = 5 )$ as well as the expected value and the variance of X are known. From this information, the following system of equations results, with which the missing probabilities $p _ { 1 } , p _ { 2 }$ and $p _ { 3 }$ can be calculated.

I $p _ { 1 } + p _ { 2 } + p _ { 3 } = 0,65$\\
II $p _ { 1 } + 2 p _ { 2 } + 3 p _ { 3 } = 1,45$\\
III $\quad 4 p _ { 1 } + p _ { 2 } = 0,6$\\
Determine, without solving the system of equations, which values for the expected value and the variance of X were used when setting up the system of equations.