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5 students of a class have an average height 150 cm and variance
18cm^2. A new student, whose height is 156cm, joined them. the variance
(in cm^2) of the height of these six students is

Show answer and solution
Let 5 students are x_1, x_2, x_3, x_4, x_5,
Given x=Σxi5=150x = \frac{\Sigma x_{i}}{5} = 150
⇒  ∑i=15\mspace2muxi=750…..(1)\sum_{i = 1}^{5}\mspace{2mu} x_{i} = 750\ldots..(1)
Σxi25−(x¯)2=18⇒\frac{\Sigma x_{i}^{2}}{5} - (\overset{¯}{x})^{2} = 18 \Rightarrow
Σxi25−(150)2=18\frac{\Sigma x_{i}^{2}}{5} - (150)^{2} = 18
⇒Σxi2=(22500+18)×5⇒∑i=15\mspace2muxi2=112590\Sigma x_{i}^{2} = (22500 + 18) \times 5 \Rightarrow \sum_{i = 1}^{5}\mspace{2mu} x_{i}^{2} = 112590........(2)
height of new student = 156 (Letx_6)
now x1+x2+x3+x4+x5+x6=750+156x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} = 750 + 156
 x¯new=x1+x2+x3+x4+x5+x66=9066=151{\overset{¯}{x}}_{new} = \frac{x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6}}{6} = \frac{906}{6} = 151........(3)
 Variance (new) =Σxi26−(x¯)2\ Variance\ (new)\ = \frac{\Sigma x_{i}^{2}}{6} - (\overset{¯}{x})^{2}

Source: Grass-G / M-28 Mathematics Paper FST 16 Jan.docx

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