a question about statistics.

Question

idku · Answer

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). `

idku · Answer

$\color{black}{\sqrt{\dfrac{(40-20)^2+R}{10}}=6}$

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

idku · Answer

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

idku · Answer

done I guess:)