Duration of sunshine(in hours) in Amritsar for first 10 days of August 1997 as reported by the Meterological Department are given as follows : 9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9
1. Find the mean $\overline{\mathrm{X}}$
2. Verify that $\sum_{i=1}^{10}\left(x^{i}-\bar{X}\right)=0$
Duration of sunshine (in hours ) for 10 days are = 9.6 , 5.2 , 3.5 , 1.5 , 1.6 , 2.4 , 2.6 , 8.4 , 10.3 , 10.9
(i) Mean $X=\frac{\text { Sum of numbers }}{\text { Total numbers }}$
$=\frac{9.6+5.2+3.5+1.5+1.6+2.4+2.6+8.4+10.3+10.9}{10}$
= 56/10 = 5.6
(ii) L.H.S $=\sum_{i=1}^{10}\left(x^{i}-\bar{X}\right)=0$
$=\left(\mathrm{x}_{1}-\overline{\mathrm{x}}\right)+\left(\mathrm{x}_{2}-\overline{\mathrm{x}}\right)+\left(\mathrm{x}_{3}-\overline{\mathrm{x}}\right)+\cdots+\left(\mathrm{x}_{10}-\overline{\mathrm{x}}\right)$
= (9.6 − 5.6) + (5.2 − 5.6) + (3.5 − 5.6) + (1.5 − 5.6) + (1.6 − 5.6) + (2.4 − 5.6) + (2.6 − 5.6) + (8.4 − 5.6) + (10.3 − 5.6) + (10.9 − 5.6)
= 4 – 0.4 – 2.1 – 4.1 – 4 – 3.2 – 3 + 2.8 + 4.7 + 5.3
= 16.8 – 16.8 = 0
= R.H.S