## idku one year ago a question about statistics.

1. idku

The list X of ten numbers has a mean of 20, and population SD, $$\color{black}{{\LARGE \sigma}_X}$$ equal to 5. THEN, Someone comes and claims that $$40$$ is the maximum on this list X. (My task is:) Show why this is impossible (i.e. why 40 can't be the maximum on the list).

2. idku

$$\color{black}{\sqrt{\dfrac{(40-20)^2+R}{10}}=6}$$ where R is the sum of all remaining component of the Standard deviation. R=($$X_2$$ - 20)$$^2$$+($$X_3$$ - 20)$$^2$$+($$X_4$$ - 20)$$^2$$+($$X_5$$ - 20)$$^2$$+($$X_6$$ - 20)$$^2$$+($$X_7$$ - 20)$$^2$$ where it is obvious that $$R\ge0$$, thus: $$\color{black}{\sqrt{\dfrac{(20)^2+R}{10}}=6}$$ $$\color{black}{\sqrt{\dfrac{400+R}{10}}=6}$$ $$\color{black}{\sqrt{\dfrac{400+R}{10}}>6}$$ knowing the restriction for R

3. idku

oh my SD is 5, not 6. But then it certainly works.

4. idku

done I guess:)