Question

An original frequency table with mean 11 and variance 9.9 was lost but the following table derived from it was found. Construct the original table.

$\begin{array}{cccccc}\text{value of deviation (d)}& -2& -1& 0& 1& 2\\ \text{Freuency(f)}& 1& 6& 7& 4& 2\end{array}$

Answer 1

Table from the given data is given below.

$\begin{array}{ccccc}d& f& d^2& fd& f{d}^2\\ -2& 1& 4& -2& 4\\ -1& 6& 1& -6& 6\\ 0& 7& 0& 0& 0\\ 1& 4& 1& 4& 4\\ 2& 2& 4& 4& 8\\ \text{Total}& \sum f=20& & \sum fd=0& \sum f{d}^2=22\end{array}$

we know $\overline{x}=a+h\frac{\sum fd}{\sum f}$

$\therefore 11=a+\frac{h\times 0}{20}\Rightarrow a=11[\because \overline{x}=11,gives]$

Also variance ${\sigma }^2=h^2[\frac{\sum f{d}^2}{\sum f}-{\left(\frac{\sum fd}{\sum f}\right)}^2]$

$[\because \sigma =99,given]$

$\therefore \left(99\right)h^2\left[\frac{22}{20}\right]-0\Rightarrow h^2=\frac{198}{22}=9$

$\Rightarrow h=3$

Mid-value would be$d=\frac{x-a}{h}\Rightarrow x=a+dh$

On putting the values of a, h and d,we get differect vlues of x,i.e.

$x=11-2\times 3$

$x=11-1\times 3$

$x=11-0\times 3$

$x=11+1\times 3$

$x=11+2\times 3$

Now all value of $x=5,8,11,14and17$

Thus original frequency table is as follows:

$\begin{array}{cccccc}\text{class}& 3.5-6.5& 6.5-9.5& 9.5-12.5& 12.5-15.5& 15.5-18.5\\ \text{Frequency}& 1& 6& 7& 4& 2\end{array}$