Answer to Question #113987 in Statistics and Probability for Vinayak

Question

The life time of a certain kind of battery is a random variable which has exponential distribution will mean of 200 hours. Find the probability that such a battery will i) last at most 100 hours i)last any where from 400 to 800 hours.

Answer 1

μ=200 so m = 1/200=0.005

1)P(x<=100)=1-e^-mx=1=1-e^-0.5=1-0.607=0.393

2)P(400<x<800)=P(x<800)-P(x<400)=(1-e^-0.005*800)-(1-e^-0.005*400)=e^-2-e^-4=0.135-0.0183=0.1167