Question

(a) A type of lightbulb is labeled as having an average lifetime of 1000 hours. It's reasonable to model the probability of failure of these bulbs by an exponential density function with mean μ = 1000. (i) Use this model to find the probability that a bulb fails within the first 500 hours. (Round your answer to three decimal places.) (ii) Use this model to find the probability that a bulb burns for more than 700 hours. (Round your answer to three decimal places.) (b) What is the median lifetime of these lightbulbs? (Round your answer to one decimal place.) hr

Answer 1

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:

$f_{X}(x)=\left \{ {{\lambda e^{-\lambda x};\ x\geq 0,\ \lambda>0} \atop {0;\ otherwise}} \right.$

The value of λ is:

$\lambda=\frac{1}{\mu}=\frac{1}{1000}=0.001$

(i)

Compute the probability that a bulb fails within the first 500 hours as follows:

$P(0\leq X\leq 500)=\int\limits^{500}_{0}{\lambda e^{-\lambda x}}\, dx$

                         $=\int\limits^{500}_{0}{0.001 e^{-0.001 x}}\, dx$

                         $=0.001\int\limits^{500}_{0}{e^{-0.001 x}}\, dx$

                         $=0.001\times |\frac{e^{-0.001x}}{0.001}|^{500}_{0}$

                         $=1-(e^{-0.001\times 500}\\=1-0.6065\\=0.3935$

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:

$P(X\geq 700)=\int\limits^{\infty}_{700}{\lambda e^{-\lambda x}}\, dx$

                   $=\int\limits^{\infty}_{700}{0.001 e^{-0.001 x}}\, dx$

                   $=0.001\int\limits^{\infty}_{700}{e^{-0.001 x}}\, dx$

                   $=0.001\times |\frac{e^{-0.001x}}{0.001}|^{\infty}_{700}$

                   $=e^{-0.001\times 700}\\=0.4966$

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:

$Median=\frac{\ln(2)}{\lambda}$

Compute the median lifetime of the lightbulbs as follows:

$Median=\frac{\ln(2)}{\lambda}$

             $=\frac{\ln(2)}{0.001}$

             $=693.15$

Thus, the median lifetime of the lightbulbs is 693.15 hours.