A certain type of light bulb has lifetimes that follows an exponential distribution with mean 1000 hours. Find the median lifetime, i.e. the lifetime x such that 50% of the light bulbs fail before x.
Solution
A certain type of light bulb has lifetimes that follows an exponential distribution with mean 100 hours.
These bulbs is distributed exponentially with a mean of $\mu = 1000$, then the probability density function of lifetime is
$$ \begin{equation*} f(t) = \frac{1}{\mu}e^{-t/\mu}\; \quad t>0. \end{equation*} $$
$$ \begin{equation*} f(t) = \frac{1}{1000}e^{-t/1000}\; \quad t>0. \end{equation*} $$
Let $M$ be the median lifetime. Then
$$ \begin{equation*} P(t < M) = 0.5 \quad \text{ OR }\quad P(t >= M) =0.5 \end{equation*} $$
$$ \begin{eqnarray*} P(t < M)= 0.5 &\Rightarrow & \int_0^M f(t) dt = 0.5 \\ &\Rightarrow & \int_0^M \frac{1}{1000}e^{-t/1000} dt = 0.5 \\ &\Rightarrow & \frac{1}{1000}\int_0^M e^{-t/1000} dt = 0.5 \\ &\Rightarrow & \frac{1}{1000}\bigg[\dfrac{-e^{-t/1000}}{1/1000}\bigg]_0^M = 0.5 \\ &\Rightarrow & \bigg[-e^{-t/1000}\bigg]_0^M = 0.5 \\ &\Rightarrow & \bigg[-e^{-M/1000}+e^{-0/1000}\bigg] = 0.5 \\ &\Rightarrow & \bigg[-e^{-M/1000}+1\bigg] = 0.5 \\ &\Rightarrow & e^{-M/1000}= 0.5 \\ &\Rightarrow & -M/1000= \log_e 0.5 \\ &\Rightarrow & -M= 1000\times (-0.69315) \\ &\Rightarrow & M= 693.15 \end{eqnarray*} $$
Thus the median lifetime of light bulb is $M=693.15$ hours
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
