I am looking at this problem and see that the variance $X_k$ calculated as follow:
$$\textrm{var}(X_k) = \frac{1}{12}\times 5^2 = \frac{25}{12}.$$
I don't quite get where that comes from. Can someone explain to me? Thanks.
2026-03-29 16:00:03.1774800003
How is the variance $X_k$ calculated in this problem?
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It's the variance of a uniform distribution. It's mentioned there that $X_k$ is uniform over the interval $[0,5]$, and therefore the variance is $$ \text{Var}[X_k] = \frac{(5-0)^2}{12} = \frac{25}{12} $$ In probability, the formula $\frac{1}{12}(b-a)^2$ is a reoccurring and important formula that should be memorised.
Note: In the context, I'm not convinced that $X_k$ actually has a uniform distribution. But in the case that we have a uniformly distributed variable, the variance is calculated as shown.