Consider the following data

The equation of straight line trend by method of least square for the above data is given by

$y=x-1999$

The correct answer is Option (3) → $y=x-1999$

$n=5,\ \sum x=10060,\ \sum y=65,\ \sum xy=130790,\ \sum x^{2}=20240730$

$b=\frac{n\sum xy - (\sum x)(\sum y)}{n\sum x^{2} - (\sum x)^{2}} =\frac{5\times130790 - 10060\times65}{5\times20240730 - (10060)^{2}}=1$

$a=\frac{\sum y - b\sum x}{n}=\frac{65 - 1\times10060}{5}=-1999$

${\,\text{Trend line: }y = 1\! \cdot x - 1999\,}$