Mean and variance of email opens

Question

A marketing campaign sends 500 emails, each opened independently with probability p = 0.12. Let O be the number of emails opened. What are the mean and variance of O?

Accepted Answer

Counting independent opened/unopened outcomes with the same open rate is binomial: O \sim 	ext{Binomial}(n=500,\ p=0.12). Use mean np and variance np(1-p): \mathbb{E}[O] = 500 	imes 0.12 = 60, \qquad \operatorname{Var}(O) = 500 	imes 0.12 	imes 0.88 = 52.8. Standard deviation \sqrt{52.8} \approx 7.27. Expect about 60 opens per send, wobbling by roughly 7. Takeaway. Binomial mean np = 60 and variance np(1-p) = 52.8; larger n makes both the mean and the spread grow.

Mean and variance of email opens

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