Question

Find the mean and standard deviation using short-cut method.
$\begin{array}{cccccccccc}{x}_i& 60& 61& 62& 63& 64& 65& 66& 67& 68\\ f_i& 2& 1& 12& 29& 25& 12& 10& 4& 5\end{array}$

Answer 1

Step 1: Finding mean
Let,
a = assumed mean be = 64
h = common factor = 1 $d_i$ = step deviation
$\begin{array}{cccccc}{x}_i& f_i& d_i=\frac{x_i-a}{h}& d_i^2& f_i{d}_i& f_i{d}_i^2\\ 60& 2& \frac{60-64}{1}=-4& 16& -4\times 2=-8& 16\times 2=32\\ 61& 1& \frac{61-64}{1}=-3& 9& -3\times 1=-3& 9\times 1=9\\ 62& 12& \frac{62-64}{1}=-2& 4& -2\times 12=-24& 12\times 4=48\\ 63& 29& \frac{63-64}{1}=-1& 1& -1\times 29=-29& 29\times 1=29\\ 64& 25& \frac{64-64}{1}=0& 0& 0\times 25=0& 25\times 0=0\\ 65& 12& \frac{65-64}{1}=1& 1& 1\times 12=12& 12\times 1=12\\ 66& 10& \frac{66-64}{1}=2& 4& 2\times 10=20& 10\times 4=40\\ 67& 4& \frac{67-64}{1}=3& 9& 3\times 4=12& 4\times 9=36\\ 68& 5& \frac{68-64}{1}=4& 16& 4\times 5=20& 5\times 16=80\\ & \sum f_i=100& & & \sum f_i{d}_i=0& \sum f_i{d}_i^2=286\end{array}$
Mean $\left(\overline{x}\right)$ = assumed mean $+\frac{\sum f_i{d}_i}{\sum f_i}\times h$
Mean $\left(\overline{x}\right)=64+\frac{0}{100}\times 1=64$
Step 2: Finding standard deviation
Standard deviation $\left(\sigma \right)=\frac{h}{\sum f_i}\sqrt{(\sum f_i).(\sum f_i{d}_i^2)-(\sum f_i{d}_i{)}^2}$
$\Rightarrow \sigma =\frac{1}{100}\sqrt{100\times 286-(0{)}^2}$
$\Rightarrow \sigma =\frac{1}{100}\sqrt{28600}$
$\Rightarrow \sigma =\frac{169.1}{100}=1.691$
hence, the mean is 64 and the standard deviation is 1.691.