Answer:
The probability that the bulb will last more than 1,200 hours is 0.3012
Step-by-step explanation:
Let the random variable X = life of a light bulb .
The random variable [tex]X\sim Exp(\lambda)[/tex]
The probability distribution function of exponential distribution is:
[tex]f(x)=\left \{ {{\lambda e^{-\lambda x};\ x>0} \atop {0};\ otherwise} \right.[/tex]
The mean of the distribution is, Mean = 1000 hours.
The value of λ is:
[tex]\lambda=\frac{1}{Mean} =\frac{1}{1000}[/tex]
Compute the probability that the bulb will last more than 1,200 hours as follows:
[tex]P(X>1200)=1-P(X\leq 1200)\\=1-\int\limits^{1200}_{0} {\lambda e^{-\lambda x}} \, dx \\=1-\lambda[\frac{e^{-\lambda x}}{-\lambda} ]^{1200}_{0}\\=1-[1-e^{-\frac{x}{1000} }]^{1200}_{0}\\=1-[1-e^{-\frac{1200}{1000}}]\\=e^{-1.2}\\=0.3012[/tex]
Thus, the probability that the bulb will last more than 1,200 hours is 0.3012.
