1. The problem statement, all variables and given/known data The expected lifetime of a battery is 0.5 and the standard deviation is 0.5. At the end of battery's lifetime, the battery will be changed so that different batteries are independent of each other. How many batteries do you need such that at least one of them is alive after 4 years with the probability 0.95? Use normal approximation. 3. The attempt at a solution I would start by calculating the number of batteries needed such that at least one of them is alive at the end of one year. I would use normal distribution, since we have the standard deviation given. Let X be the number of batteries such that at least one of the batteries is alive at the end of one year X ~ N(upperbound, expected value of batteries that break in one year, standard deviation of batteries) I think that the above values are the following in the simplified situation. X ~N(1, 0.5n, 0.5n) The standard deviation is got by st(X) = 0.5 for one battery st(X) = 0.5n for n number of batteries I also think that the expected value of batteries that break is 0.5n. For example, I get the probability, 0.841 (= normalcdf(-E99, 1, 0.5, 0.5)), that one battery is alive at the end of one year. This result may be wrong, since the expected value is 0.5 that one battery is alive at the end of one year. This suggests me that the use of normal distribution may be a wrong choice at least at the beginning of the problem. Perhaps, we need to use binomial distribution first for one battery, and then apply normal approximation to get the number of batteries for 4 years. Please, let me know how you would solve the problem.