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 300 hours. (Round your answer to three decimal places.) 0.259 Correct: Your answer is correct. (ii) Use this model to find the probability that a bulb burns for more than 900 hours. (Round your answer to three decimal places.) 0.407 Correct: Your answer is correct. (b) What is the median lifetime of these lightbulbs? (Round your answer to one decimal place.) 693 Incorrect: Your answer is incorrect. hr

Answer 1

a)0.259

b)0.407

c)693.1hrs

Explanation:

This is an exponential density function with a constant c and mean of 1000:

`f(t)={0,t<0\choose ce^(-ct,t\geq )0`

The constant c is calculated as:

`\mu=(1)/(c)\\\\c=(1)/(\mu)=(1)/(1000)\\\\c=0.001`

#To find the probability that i falls with the first 300 hrs;

`P(0\leq X\leq 300)=\int\limits^(300)_0 {f(t)} \, dt\\\\=\int\limits^(300)_0{0.001e^(-0.001t)dt\\\\\\`

`=[-e^(-0.001t)]\limits^(300)_(0)\\\\=-0.7408+1\\\\=0.2592`

Hence, the probability that a bulb fails within the first 300 hours is 0.259

b. For more than 900hrs, integrate between 900 and infinity.

#Use the limit method for dealing with improper integrals:

`P(X\geq 900)=\int\limits^(\infty)_(900) {f(t)} \, dt\\ \\\\=\int\limits^(\infty)_(900){0.001e^(-0.001t)dt\\\\`

`=\int\limits^(b)_(900){0.001e^(-0.001t)dt, b->\infty\\\\`

`=[-e^(-0.001t)]\limits^(b)_(900)\\\\=[-e^(-0.001t)]\limits^(b)_(900)\\\\=0--0.4066\\\\=0.407`

Hence, the probability that a bulb burns for more than 900 hours is 0.407

c. To find the median lifetime of these bulbs:

`P(X\leq median)=P(X\geq median)=0.5\\\\P(X\geq median)=e^(-median/\mu)=0.5\\\\-median/1000=In(0.5)\\\\-median=1000\ In(0.5)\\\\median=693.1`

Hence, the median lifetime of these bulbs is 693.1hrs