A bank accepts rolls of pennies and gives 50 cents credit to a customer without counting the contents. Assume that a roll contains 49 pennies 30 percent of the time, 50 pennies 60 percent of the time, and 51 pennies 10 percent of the time. How many rolls does the bank need to collect to have a 99 percent chance of a net loss?

• 2,400
• 2,435
• 3,450
• 4,500
• 4,870

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

Solution: E

The expected number of pennies in a single roll equals [math]\mu = 49.8 [/math] and the variance equals [math]\sigma^2 = 0.36 [/math]. In particular, the net loss for [math]n[/math] rolls is approximately normally distributed with mean [math]0.02n[/math] and variance [math]n\sigma^2[/math]. Hence the probability of a net loss equals

[[math]]P(Z \geq \frac{-0.02n}{\sigma \sqrt{n}}) = P(Z \geq \frac{-\sqrt{n}}{30})[[/math]]

where [math]Z[/math] is a standard normal variable. The 1^th percentile for a standard normal equals -2.326, hence we need

[[math]] \frac{\sqrt{n}}{30} \geq 2.326 \implies n \geq 4869.25 [[/math]]