Revision as of 00:32, 25 June 2024 by Admin (Created page with "'''Solution: B''' We have <math display = "block">P(X=i) = \frac{i}{\sum_{i=1}^6 i } = \frac{i}{21}</math>, <math display = "block"> E[X] = \frac{1}{21}\sum_{i=1}^6 i^2 = \frac{91}{21} </math> and <math display = "block"> \operatorname{E}[X^2] = \frac{1}{21}\sum_{i=1}^6 i^3 = 21. </math> Hence the variance of <math>X</math> equals 21 - (91/21)<sup>2</sup> = 2.22.")
Exercise
ABy Admin
Jun 25'24
Answer
Solution: B
We have
[[math]]P(X=i) = \frac{i}{\sum_{i=1}^6 i } = \frac{i}{21}[[/math]]
,
[[math]]
E[X] = \frac{1}{21}\sum_{i=1}^6 i^2 = \frac{91}{21}
[[/math]]
and
[[math]]
\operatorname{E}[X^2] = \frac{1}{21}\sum_{i=1}^6 i^3 = 21.
[[/math]]
Hence the variance of [math]X[/math] equals 21 - (91/21)2 = 2.22.