Suppose that in a particular application requiring a single battery, the mean lifetime of the battery is 4 weeks, with a standard deviation of 1 week. The battery is replaced by a new one when it dies and so on. Assume lifetimes of batteries are independent. What, approximately, is the probability that more than 26 replacements will have to be made in a two-year period, starting at the time of installation of a new battery, and not counting that new battery as replacement? [Hint: Use the normal approximation to the distribution of the total lifetime of $n$ batteries for a suitable $n$].
$X$ = lifetime of a battery
$E(X) = 4$
$SD(X) = 1$
Let $S =$ total lifetime of $n$ replacement batteries used in a 2-year period.
$P(S > 26X)$
$= P(\frac{S-E(S)}{SD(S)} > \frac{26X-E(S)}{SD(S)})$
$= P(z > \frac{26X-E(S)}{SD(S)})$, where $z = \frac{S-E(S)}{SD(S)}$
$= P(z > \frac{26X-4n}{\sqrt{Var(S)}})$
$= P(z > \frac{26X-4n}{\sqrt{n}})$
$=$ .....?
Textbook Answer:
Approximately $\Phi(-0.77) \approx 0.22$
Edit 1:
Let $S_n =$ total lifetime of $n$ batteries.
$2$ years $= 52 \cdot 2 = 104$ weeks
$P(S_{1+26} < 104)$, if the new battery + 26 replacements have a lifetime of less than 2 years, that means we'll need more than 26 replacements in that 2 year.
$= P(S_{27} < 104)$
$= P(\frac{S_{27}-E(S_{27})}{SD(S_{27})} < \frac{104-E(S_{27})}{SD(S_{27})})$
$= P(z < \frac{104-E(S_{27})}{SD(S_{27})})$, where $z =\frac{S_{27}-E(S_{27})}{SD(S_{27})}$
$= P(z < \frac{104-108}{\sqrt{\sum_{1}^{27}Var(X_i)}})$
$= P(z < \frac{104-108}{\sqrt{\sum_{1}^{27}1^2}})$
$= P(z < \frac{104-108}{\sqrt{27}})$
$= P(z < -0.769800358)$
$= \Phi(-0.769800358)$
$= 1 - \Phi(0.769800358)$
$= 1 - 0.7764$
$= 0.2236$
We want to know the probability that the sum of 27 lifespans is less than 2 years (since the event of interest is that more than 26 replacements will have been made durring those 104 weeks). $$\begin{array} P(S_{27}<104) &= P(Z < (104-27*4)/\sqrt{27*1^2})\\[1ex]& = \Phi(-4\sqrt{3}/9)\\[1ex] &\approx \Phi (-0.7698) \end{array}$$