Given,
N = 2a,
Mean,
\(\bar{x}=\frac{a+a+...+a+(-a)+(-a)+...(-a)}{2n}\)
\(=\frac{0}{2n}\)
= 0
Standard deviation,
\(σ=2\)
⇒ \(σ^2=2^2\)
\(⇒\frac{\displaystyle\sum {x_i}^2}{N}-(\frac{\displaystyle\sum {x_i}}{N})^2=4\)
\(⇒\frac{\displaystyle\sum {x_i}^2}{N}-x^{-2}=4\)
\(⇒\frac{\displaystyle\sum {x_i}^2}{N}-0=4\)
\(⇒\displaystyle\sum {x_i}^2=8n\)
⇒ a2 + a2 + ⋯ + a2 (to 2n terms) 2a2 = 8n
⇒ a2 = 4
⇒ a = ± 2
∴ |a| = |± 2|
= 2