There are $60$ students in a class. The following is the frequency distribution of the marks obtained by the students in a test:
$\begin{array}{|l|l|l|l|l|l|l|} \hline \text{Marks} & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline \text{Frequency} & x-2 & x & x^2 & (x+1)^2 & 2x & x+1 \\ \hline \end{array}$
where $x$ is a positive integer. Determine the mean and standard deviation of the marks.

(N/A) Sum of frequencies:
$(x-2) + x + x^2 + (x+1)^2 + 2x + (x+1) = 60$
$x^2 - 2 + x + x^2 + x^2 + 2x + 1 + 2x + x + 1 = 60$
$3x^2 + 7x - 60 = 0$
$(3x - 12)(x + 5) = 0$ (Wait,re-calculating: $x-2+x+x^2+x^2+2x+1+2x+x+1 = 2x^2+7x = 60 \implies 2x^2+7x-60=0$)
$(2x+15)(x-4)=0$
Since $x$ is a positive integer,$x=4$.
Frequency table for $x=4$:
$\begin{array}{|c|c|c|c|c|} \hline x_i & f_i & d_i=x_i-3 & f_i d_i & f_i d_i^2 \\ \hline 0 & 2 & -3 & -6 & 18 \\ \hline 1 & 4 & -2 & -8 & 16 \\ \hline 2 & 16 & -1 & -16 & 16 \\ \hline 3 & 25 & 0 & 0 & 0 \\ \hline 4 & 8 & 1 & 8 & 8 \\ \hline 5 & 5 & 2 & 10 & 20 \\ \hline \text{Total} & \Sigma f_i=60 & & \Sigma f_i d_i=-12 & \Sigma f_i d_i^2=78 \\ \hline \end{array}$
Mean $= A + \frac{\Sigma f_i d_i}{\Sigma f_i} = 3 + (\frac{-12}{60}) = 3 - 0.2 = 2.8$
Standard Deviation $\sigma = \sqrt{\frac{\Sigma f_i d_i^2}{\Sigma f_i} - (\frac{\Sigma f_i d_i}{\Sigma f_i})^2} = \sqrt{\frac{78}{60} - (-0.2)^2} = \sqrt{1.3 - 0.04} = \sqrt{1.26} \approx 1.12$