The probability distribution of a random variable is given below:

Then $\mathrm{P}\left(0<\mathrm{X}<5\right)$

Since $\sum \mathrm{P}(\mathrm{X}=\mathrm{x})=1$ then

$$\begin{gathered} \Rightarrow k+2k+2k+3k+k^2+2k^2+7k^2+k=1 \\ \Rightarrow 10k^2+9k-1=0 \\ \Rightarrow 10k^2+10k-k-1=0 \\ \Rightarrow 10k (k+1)-1(k+1)=0 \\ \Rightarrow (k+1) (10k-1)=0 \\ \Rightarrow =-1 \left(\mathrm{or}\right) k=\frac{1}{10} \Rightarrow k=\frac{1}{10} (\because 0\le p\le 1) \end{gathered}$$

Now 

 $\begin{array}{rcl}\mathrm{P}(0<X& <& 5)=\mathrm{P}(X=1)+\mathrm{P}(X=2)+\mathrm{P}(X=3)+\mathrm{P}(X=4)\\ & =& k+2k+2k+3k\\ & =& 8k\\ & =& 8\left(\frac{1}{10}\right)\\ & =& \frac{8}{10}\end{array}$