Given: Incorrect mean, x=12 and incorrect standard deviation, σ′=3.
⇒15Σxi=12
⇒Σxi=180
Correct mean is given by,
μ=15Σxi−10+12
⇒μ=15182
Also, σ′=15∑xi2−144=3
⇒15∑xi2=9+144
⇒∑xi2=2295
Correct Σxi2=2295−100+144=2339
σ2 (correct variance) =152339−15×15182×182
⇒15(μ+μ2+σ2)=15(15182+15×15182×182+152339−15×15182×182)
⇒15(μ+μ2+σ2)=15(15182+152339)
⇒15(μ+μ2+σ2)=2521