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Math Help - Normal Distribution life time of batteries

  1. #1
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    Normal Distribution life time of batteries

    Suppose that the amount of time that a certain battery functions is a normal random variable with mean 300 hours and standard deviation 60 hours. suppose that an individual owns three such batteries, two of which are spares. the lifetimes of the batteries are independent.

    i) what is the probability that the total liftimes of the three batteries exceeds 1000 hours?

    Is this the correct way? Y has N(900,9*60^{2})
    do i find P(Y>1000)=P(Z>(1000-900)/{180})

    ii) what is the probability that the lifetimes of the first battery is more than the sum of the lifetimes of the other two batteries?

    I don't know how to start this so any help would be greatly appreciated.

    Thanks
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  2. #2
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    Quote Originally Posted by charikaar View Post
    Suppose that the amount of time that a certain battery functions is a normal random variable with mean 300 hours and standard deviation 60 hours. suppose that an individual owns three such batteries, two of which are spares. the lifetimes of the batteries are independent.

    i) what is the probability that the total liftimes of the three batteries exceeds 1000 hours?

    Is this the correct way? Y has N(900,9*60^{2})
    do i find P(Y>1000)=P(Z>(1000-900)/{180})

    ii) what is the probability that the lifetimes of the first battery is more than the sum of the lifetimes of the other two batteries?

    I don't know how to start this so any help would be greatly appreciated.

    Thanks
    U = X_1 + X_2 + X_3 ~ Normal \left( \mu = 300 + 300 + 300, \, \sigma^2 = 60^2 + 60^2 + 60^2\right).

    W = X_1 - (X_2 + X_3) ~ Normal \left( \mu = 300 - 300 - 300, \, \sigma^2 = 60^2 + 60^2 + 60^2\right).
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