A company manufactures a brand of light bulb with a lifetime in months that is normally distributed with mean 3 and variance 1. A consumer buys a number of these bulbs with the intention of replacing them successively as they burn out. The light bulbs have mutually independent lifetimes.

Calculate the smallest number of bulbs to be purchased so that the succession of light bulbs produces light for at least 40 months with probability at least 0.9772.

• 14
• 16
• 20
• 40
• 55

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Solution: B

Let [math]X_1, \ldots, X_n [/math] denote the life spans of the [math]n[/math] light bulbs purchased. Since these random variables are independent and normally distributed with mean 3 and variance 1, the random variable [math]S = X_1 + \cdots + X_n [/math] is also normally distributed with mean [math]\mu = 3n [/math] and standard deviation [math] \sigma = \sqrt{n}[/math]. Now we want to choose the smallest value for [math]n[/math] such that

[[math]] 0.9772 = \operatorname{P}[S \gt 40] = \operatorname{P}[ \frac{S-3n}{\sqrt{n}} \gt \frac{40-3n}{\sqrt{n}}]. [[/math]]

This implies that [math]n[/math] should satisfy the following inequality:

[[math]] -2 \geq \frac{40-3n}{\sqrt{n}}. [[/math]]

To find such an [math]n[/math], let’s solve the corresponding equation for [math]n[/math]:

[[math]] \begin{align*} -2 = \frac{40-3n}{\sqrt{n}} \\ -2\sqrt{n} = 40 -3n \\ 3n-2\sqrt{n} -40 = 0 \\ (3\sqrt{n}+10)(\sqrt{n} -4) = 0 \\ \sqrt{n} = 4 \\ n = 16 \end{align*} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.