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

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

$\begin{tabular}{|l|l|l|l|l|l|l|} \hline Marks & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline Frequency & $x-2$ & $x$ & $x^{2}$ & $(x+1)^{2}$ & $2 x$ & $x+1$ \\ \hline \end{tabular}$

Where $x$ is a positive integer. Determine the mean and standard deviation of the marks.

CBSE | Class 11 | Maths | NCERT Exemplar | Statistics

There are 60 students in a class. The following is the frequency distribution of the marks obtained by the students in a test:

$\begin{tabular}{|l|l|l|l|l|l|l|} \hline Marks & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline Frequency & $x-2$ & $x$ & $x^{2}$ & $(x+1)^{2}$ & $2 x$ & $x+1$ \\ \hline \end{tabular}$

Where $x$ is a positive integer. Determine the mean and standard deviation of the marks.

Solution:

It is given that there are 60 students in a class. Also the frequency distribution of the marks obtained by the students in a test is given.

We now need to find the mean and the standard deviation of the marks.

Given that there are 60 students in the class, so

\section{ $\sum \mathrm{f}_{\mathrm{i}}=60$ }

$\Rightarrow(x-2)+x+x^{2}+(x+1)^{2}+2 x+x+1=60$

$\Rightarrow 5 x-1+x^{2}+x^{2}+2 x+1=60$

$\Rightarrow 2 x^{2}+7 x=60$

$\Rightarrow 2 x^{2}+7 x-60=0$

Splitting the middle term, we get

$\Rightarrow 2 x^{2}+15 x-8 x-60=0$

$\Rightarrow x(2 x+15)-4(2 x+15)=0$

$\Rightarrow(2 x+15)(x-4)=0$

$\Rightarrow 2 x+15=0$ or $x-4=0$

$\Rightarrow 2 x=-15$ or $x=4$

Given $x$ is a positive number, so $x$ can take 4 as the only value.

And let’s say the assumed mean, $a=3$ .

Putting $x=4$ and $a=3$ in the frequency distribution table and add other columns after calculations, we obtain

$\begin{tabular}{|l|l|l|l|l|} \hline Marks $\left(x_{i}\right)$ & Frequency $\left(f_{i}\right)$ & $d_{i}=x_{i}-a$ & $f_{i} d_{i}$ & $f_{i} d_{i}^{2}$ \\ \hline 0 & $x-2=4-2=2$ & $-3$ & $-6$ & 18 \\ \hline 1 & $x=4$ & $-2$ & $-8$ & 16 \\ \hline 2 & $x^{2}=4^{2}=16$ & $-1$ & $-16$ & 16 \\ \hline 3 & $(x+1)^{2}=25$ & 0 & 0 & 0 \\ \hline 4 & $2 x=8$ & 1 & 4 & 8 \\ \hline 5 & $x+1=5$ & 2 & 10 & 20 \\ \hline Total & 60 & & $-12$ & 78 \\ \hline \end{tabular}$

And it is known that the standard deviation is

$\sigma=\sqrt{\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}^{2}}{\mathrm{n}}-\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{d}_{\mathrm{i}}}{\mathrm{n}}\right)^{2}}$

Substitute values from above table,

$\begin{aligned} &\sigma=\sqrt{\frac{78}{60}-\left(\frac{-12}{60}\right)^{2}} \\ &\sigma=\sqrt{1.3-(0.2)^{2}} \\ &\sigma=\sqrt{1.3-0.04} \\ &\Rightarrow \sigma=1.12 \end{aligned}$