In a quiet hour, which has probability 1/2, the number of fills is Poisson with mean 2. In a busy hour, with probability 1/2, it is Poisson with mean 6. What is the variance of the number of fills in a randomly chosen hour?

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Fills in a random hour

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In a quiet hour, which has probability $\frac{1}{2}$ , the number of fills is Poisson with mean $2$ . In a busy hour, with probability $\frac{1}{2}$ , it is Poisson with mean $6$ . What is the variance of the number of fills in a randomly chosen hour?

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Hint

A Poisson count has variance equal to its mean; then apply the total variance decomposition.

Solution

The average within-hour variance equals the average mean, $\frac{2 + 6}{2} = 4$ .
The means $2$ and $6$ are equally likely, so their variance is $2^{2} = 4$ .
The total variance is $4 + 4 = 8$ .

Answer: $8$

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