Question

A person has 100 light bulbs whose lifetimes are independent exponentials with mean 5 hours. If the bulbs are used one at a time, with a failed bulb being replaced immediately by a new one, approximate the probability that there is still a working bulb after 525 hours.

Short answer

$P\{S_{100}>525\}=0.3085$

Step-by-step solution

Step 1. Given information

Let X_ibe a random variable that represents the lifetime of the i^thbulb, i = 1,...,100.

Step 2. Mean and variance of Xi

X_iare i.i.d.'s all having exponential distribution mean 5 hours and variance 25 hours.

$$\begin{gathered} E\left(X_i\right)=5hours \\ V\left(X_i\right)=5^2=25hours \end{gathered}$$

Step 3. Defining S100

$S_{100}=\underset{i=1}{\overset{100}{\sum X_i}}$

Step 4. To find

$P\{S_{100}>525\}$

Step 5. By using central limit theorem

$$\begin{gathered} P\{S_{100}>525\}=P\left\{\frac{S_{100}-100\times 5}{\sqrt{100\times 25}}>\frac{525-100\times 5}{\sqrt{100\times 25}}\right\} \\ P\{S_{100}>525\}=P\left\{\frac{S_{100}-500}{\sqrt{2500}}>\frac{525-500}{\sqrt{2500}}\right\} \\ P\{S_{100}>525\}=P\left\{Z>\frac{25}{50}\right\} \\ P\{S_{100}>525\}=P\left\{Z>0.5\right\} \\ P\{S_{100}>525\}=1-P\{Z<0.5\}=1-0.6915 \\ P\{S_{100}>525\}=0.3085 \end{gathered}$$

Step 6. Final answer

The probability that there is still a working bulb after 525 hours is 0.3085