Let x and y be the year and students strength.
Here X̄ = middle most year = 2012
From the normal equation: na + b Σx + c Σx2 = Ey
5a + b.0 + c.10 = 53; 5a + 10c = 53 ………. (1)
From: a Σx + b Σx2 + c Σx3 – Σxy; a.0 + b.10 + c.0 = 0
10 b = 0
∴ b = 0.
And from: a Σx2 + b Σx3 + c Σx4 = Σx2y; a. 10 + b.0 + c.34 = 120
i. e., 10a + 34c = 120 …….(2)
From (1) and (2)
∴ c = 1. put c = 1 in (1) we get 5a +10 (1) = 53, 5a = 53 – 10
a = \(\frac{43}{5}\) = 8.6.
The fitted parabolic trend equation is: ŷ = a + bx + cx2
ŷ = 8.6 + 0x + 1x2 ; ŷ = 8.6 + x2.