Given,
N = 10,
σ = 16, and
\(\bar{x}\) = 12.
Let the new mean be \(\bar{x_1}\) and new variance be σ12.
Here,
4\(\bar{x}\) = 4\(\bar{x}\)
We Know,
\(σ_1^2=\frac{\sum {x'}_i^2}{N}-(\frac{\sum \bar{x}'}{N})^2\)
\(=\frac{\sum (4x_i)^2}{N}-(\frac{\sum \bar{4x_i}}{N})^2\)
\(=16\{\frac{\sum x_i^2}{N}-(\frac{\sum {x_i}}{N})^2\}\)
= 16 × σ2
= 16 × 16
= 256
Also,
\(\bar{x_1}=\frac{\sum x_i}{N}\)
\(=4\frac{\sum x_i}{N}\)
\(=4\bar{x}\)
\(=\frac{\sum 4x_i}{N}\)
= 4 x 12
= 48
∴ New mean = 48 and new variance = 256.