Answer: 0.5507
Step-by-step explanation:
The cumulative distribution function for exponential distribution:-
[tex]P(X\leq x)=1-e^{-\lambda x}[/tex]
Given : The life of a light bulb is exponentially distributed with a mean of 1,000 hours.
Then , [tex]\lambda=\dfrac{1}{1000}[/tex]
Then , the probability that the bulb will last less than 800 hours is given by :-
[tex]P(X\leq 800)=1-e^{\frac{-800}{1000}}\\\\=1-e^{-0.8}=0.5506710358\approx0.5507[/tex]
Hence, the probability that the bulb will last less than 800 hours = 0.5507
The probability that the bulb will last less than 800 hours is 0.550
An incandescent light bulb is an electric light with a wire filament heated to such a high temperature that it glows with visible light. The filament of an incandescent light bulb is protected from oxidation with a glass or fused quartz bulb that is filled with inert gas or a vacuum. The exponential distribution is the probability distribution of the time between events in a Poisson point process. It is a particular case of the gamma distribution. While the arithmetic mean is the central value of a discrete set of numbers: specifically, the sum of the values divided by the number of values
The life of a light bulb is exponentially distributed with a mean of 1,000 hours. What is the probability that the bulb will last less than 800 hours?
What is the probability that the light bulb will have to replaced within 800 hours?
[tex]\lambda = \frac{1}{1000}[/tex]
[tex]P(x<=800) = 1 - e^{(-\lambda*x)} = 1-e^{[(-\frac{1}{1000} )*800]} = 1-e^{-0.8} = 1-0.449 = 0.550[/tex]
Grade: 5
Subject: math
Chapter: probability
Keywords: mean, bulb, hours, exponential, light
