Respuesta :
Answer:
(a)
(i) The probability that a bulb fails within the first 500 hours is 0.3935.
(ii) The probability that a bulb burns for more than 700 hours is 0.4966.
(b) The median lifetime of the lightbulbs is 693.15 hours.
Step-by-step explanation:
Let X = lifetime of a type of lightbulb.
The random variable X is Exponentially distributed with mean lifetime of, μ = 1000 hours.
The probability density function of X is:
[tex]f_{X}(x)=\left \{ {{\lambda e^{-\lambda x};\ x\geq 0,\ \lambda>0} \atop {0;\ otherwise}} \right.[/tex]
The value of λ is:
[tex]\lambda=\frac{1}{\mu}=\frac{1}{1000}=0.001[/tex]
(i)
Compute the probability that a bulb fails within the first 500 hours as follows:
[tex]P(0\leq X\leq 500)=\int\limits^{500}_{0}{\lambda e^{-\lambda x}}\, dx\\[/tex]
[tex]=\int\limits^{500}_{0}{0.001 e^{-0.001 x}}\, dx\\[/tex]
[tex]=0.001\int\limits^{500}_{0}{e^{-0.001 x}}\, dx\\[/tex]
[tex]=0.001\times |\frac{e^{-0.001x}}{0.001}|^{500}_{0}[/tex]
[tex]=1-(e^{-0.001\times 500}\\=1-0.6065\\=0.3935[/tex]
Thus, the probability that a bulb fails within the first 500 hours is 0.3935.
(ii)
Compute the probability that a bulb burns for more than 700 hours as follows:
[tex]P(X\geq 700)=\int\limits^{\infty}_{700}{\lambda e^{-\lambda x}}\, dx\\[/tex]
[tex]=\int\limits^{\infty}_{700}{0.001 e^{-0.001 x}}\, dx\\[/tex]
[tex]=0.001\int\limits^{\infty}_{700}{e^{-0.001 x}}\, dx\\[/tex]
[tex]=0.001\times |\frac{e^{-0.001x}}{0.001}|^{\infty}_{700}[/tex]
[tex]=e^{-0.001\times 700}\\=0.4966[/tex]
Thus, the probability that a bulb burns for more than 700 hours is 0.4966.
(b)
The median of an exponentially distributed random variable is:
[tex]Median=\frac{\ln(2)}{\lambda}[/tex]
Compute the median lifetime of the lightbulbs as follows:
[tex]Median=\frac{\ln(2)}{\lambda}[/tex]
[tex]=\frac{\ln(2)}{0.001}[/tex]
[tex]=693.15[/tex]
Thus, the median lifetime of the lightbulbs is 693.15 hours.
