Question

The mean of $15$ observations is $32$. Find the resulting mean, if each observation is decreased by $20\%$

• A. $38.4$
• B. $40.5$
• C. $37$
• D. None of these

Answer 1

The correct option is D None of these

$Letthe15observationsbe{x}_1,x_2,x_3,.........,x_{13},x_{14},x_{15}GivenMeanofobservations=32\Rightarrow \frac{\sum (x_1+x_2+x_3+.......x_{13}+x_{14}+x_{15})}{15}=32\Rightarrow \sum (x_1+x_2+x_3+.......x_{13}+x_{14}+x_{15})=32\times 15=480Wheneachobservationisdecreasedby20percent.Thensumofobservations=\sum (x_1-x_1\times \frac{20}{100})+(x_2-x_2\times \frac{20}{100})+(x_3-x_3\times \frac{20}{100})+...........+(x_{13}-x_{13}\times \frac{20}{100})+(x_{14}-x_{14}\times \frac{20}{100})+(x_{15}-x_{15}\times \frac{20}{100})=\sum (x_1+x_2+x_3+.......x_{13}+x_{14}+x_{15})-\sum (x_1\times \frac{20}{100}+x_2\times \frac{20}{100}+x_3\times \frac{20}{100}+...........+x_{13}\times \frac{20}{100}+x_{14}\times \frac{20}{100}+x_{15}\times \frac{20}{100})=480-\sum [(x_1+x_2+x_3+.......x_{13}+x_{14}+x_{15})\times \frac{20}{100}]=480-\frac{20}{100}\sum (x_1+x_2+x_3+.......x_{13}+x_{14}+x_{15})=480-\frac{20}{100}\times 480=480(1-0.2)=480\times 0.8=384NewMean=\frac{Sumofobservations}{Numberofobservations}=\frac{384}{15}=25.6$