A lamp has two bulbs, each of a type with average lifetime 1400 hours. Assuming that we can model the probability of failure of these bulbs by an exponential density function with mean μ = 1400, find the probability that both of the lamp's bulbs fail within 1500 hours. (Round your answer to four decimal places.)

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Answer:

The probability of failure of both the bulbs  is 0.4323.

Step-by-step explanation:

For an exponential distribution the distribution is given by

[tex]f(x,\lambda )=\int_{0}^{x }\lambda e^{-\lambda x}dx[/tex]

The value of λ is related to the mean μ as λ=1/μ,

Let us denote the 2 bulbs by X and Y thus the probability distribution of the 2 bulbs is as under

[tex]P(X)=\int_{0}^{x }\lambda _{X}e^{-\lambda _{X}x}dx[/tex]

Similarly for the bulb Y the distribution function is given by

[tex]P(Y)=\int_{0}^{y }\lambda _{Y}e^{-\lambda _{Y}y}dy[/tex]

Thus the probability for both the bulbs to fail within 1500 hours is

[tex]P(E)=\int_{0}^{1500}\int_{0}^{1500}\frac{1}{1400}e^{\frac{-x}{1400}}\cdot \frac{1}{1400}e^{\frac{-y}{1400}}dxdy\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}\int_{0}^{1500}e^{\frac{-x}{1400}}\cdot e^{\frac{-y}{1400}}dxdy)\\\\P(E)=\frac{1}{1400^2}(\int_{0}^{1500}e^{\frac{-x}{1400}}dx)\cdot (\int_{0}^{1500}e^{\frac{-y}{1400}}dy)\\\\P(E)=\frac{1}{1400^{2}}\times 920.473\times 920.473\\\\\therefore P(E)=0.4323[/tex]

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