The lifetime of a lightbulb in a certain application is normally distributed with mean μ = 1500 hours and standard deviation σ = 200 hours. NOTE: This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. What is the probability that a lightbulb will last more than 1800 hours?

Answer :

Answer:

0.000647588

Step-by-step explanation:

The normal probability function formula is:

[tex]f(x)={\frac {1}{\sigma {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {x-\mu }{\sigma }}\right)^{2}}}[/tex]

To find the probability that a lightbulb will last more than 1800 hours, we can plug in x = 1800, μ = 1500, and σ = 200

[tex]f(1800) = {\frac {1}{200 {\sqrt {2\pi }}}}e^{-{\frac {1}{2}}\left({\frac {1800-1500}{200 }}\right)^{2}}}[/tex]

[tex]f(1800) = {\frac {1}{501.3}e^{-1.125}}[/tex]

[tex]f(1800) = 0.000647588[/tex]

So the probability that a lightbulb will last more than 1800 hours is 0.000647588

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