If for some \(x \in \mathbb{R}^{+} \cup\{0\}\), the frequency distribution of the marks obtained by 20 students in a test is,

then the mean of the marks is

\(\begin{aligned}
& \text {Here, } \Sigma \mathrm{f}_{\mathrm{i}}=(x+1)^2+2 x-5+x^2-3 x+x \\
& =2 x^2+2 x-4 \\
& \sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}=2(x+1)^2+3(2 x-5)+5\left(x^2-3 x\right)+7 x \\
& =7 x^2+2 x-13
\end{aligned}\)
\(\begin{aligned}
& \mathrm{N}=20 \\
& \Rightarrow \sum \mathrm{f}_{\mathrm{i}}=20 \\
& \Rightarrow 2 x^2+2 x-4=20 \\
& \Rightarrow x=-4,3 \\
& \Rightarrow x=3
\end{aligned}\)
\(\ldots\left[\because x \in \mathrm{R}^{+} \cup\{0\}\right]\)
Now mean
\(\begin{aligned}
(\bar{x}) & =\frac{\sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}}{\mathrm{~N}} \\
& =\frac{7(3)^2+2(3)-13}{20}
\end{aligned}\)
\(=2.8\)