Question

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

$\begin{array}{cc}\text{Marks}& \text{Frequency}\\ 0& x-2\\ 1& x\\ 2& x^2\\ 3& (x+1{)}^2\\ 4& 2x\\ 5& x+1\end{array}$

Where $x$ is positive integer. Then the variance of the marks is

• A. $1.26$
• B. $1.28$
• C. $1.34$
• D. $1.24$

Answer 1

The correct option is A $1.26$

Sum of frequencies
$\sum f_i=60$

$x-2+x+x^2+(x+1{)}^2+2x+x+1=60\Rightarrow 2x^2+7x-60=0\Rightarrow (x-4\left)\right(2x+15)=0\therefore x=4(\text{As}x-2\text{is frequency, so}x-2\ge 0)$

Now,
$\begin{array}{cccc}{x}_i& f_i& f_i{x}_i& f_i{x}_i^2\\ 0& 2& 0& 0\\ 1& 4& 4& 4\\ 2& 16& 32& 64\\ 3& 25& 75& 225\\ 4& 8& 32& 128\\ 5& 5& 25& 125\\ & & \sum f_i{x}_i=168& \sum f_i{x}_i^2=546\end{array}$

$\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}=\frac{168}{60}=2.8$

${\sigma }^2=\frac{1}{N}\sum f_i{x}_i^2-(\overline{x}{)}^2$
$=\frac{546}{60}-(2.8{)}^2=1.26$