If both the mean and the standard deviation
Question:

If both the mean and the standard deviation of 50 observations $\mathrm{x}_{1}, \mathrm{x}_{2} \ldots, \mathrm{x}_{50}$ are equal to 16 , then the mean of $\left(x_{1}-4\right)^{2},\left(x_{2}-4\right)^{2}, \ldots . .\left(x_{50}-4\right)^{2}$ is :

  1. 525 

  2. 380

  3. 480

  4. 400


Correct Option: , 4

Solution:

$\operatorname{Mean}(\mu)=\frac{\sum \mathrm{x}_{\mathrm{i}}}{50}=16$

standard deviation $(\sigma)=\sqrt{\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{50}-(\mu)^{2}}=16$

$\Rightarrow(256) \times 2=\frac{\sum x_{i}^{2}}{50}$

$\Rightarrow$ New mean

$=\frac{\sum\left(x_{i}-4\right)^{2}}{50}=\frac{\sum x_{i}^{2}+16 \times 50-8 \sum x_{i}}{50}$

$=(256) \times 2+16-8 \times 16=400$

Administrator

Leave a comment

Please enter comment.
Please enter your name.