Geologists estimate the time since the most recent cooling of a mineral by counting the number of uranium fission tracks on the surface of t

Question

Geologists estimate the time since the most recent cooling of a mineral by counting the number of uranium fission tracks on the surface of the mineral. A certain mineral specimen is of such an age that there should be an average of 6 tracks per cm^2 of surface area. Assume the number of tracks in an area follows a Poisson distribution. Let X represent the number of tracks counted in 1 cm^2 of surface area. Find:(a) P(X = 7)(b) P(X ≥ 3)(c) P(2 < X < 7)(d) μX(e) σX

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Mackenzie 3 weeks 2021-09-26T04:22:26+00:00 1 Answer 0

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    2021-09-26T04:23:41+00:00

    Answer:

    a) 0.138

    b) 0.938

    c) 0.544

    d) 6

    e) 2.45

    Step-by-step explanation:

    Let X represent the number of tracks counted in 1 cm^2 of surface area.

    The estimated average is 6 tracks per cm^2. This is the parameter λ of the Poisson distribution.

    The expression for the Poisson distribution is

    P(x=k)=\frac{\lambda^ke^{-\lambda}}{k!}

    Then

    a) x=7

    P(x=7)=\frac{\lambda^7e^{-\lambda}}{7!}=\frac{6^7e^{-6}}{7!}=\frac{693.89}{5040}=0.138

    b) x≥3

    P(x\geq 3)=1-(P(0)+P(1)+P(2))\\\\P(0)=\frac{6^0e^{-6}}{0!}=\frac{0.00248}{1}  =0.00248\\\\P(1)=\frac{6^1e^{-6}}{1!}=\frac{0.01487}{1}  =0.01487\\\\P(2)=\frac{6^2e^{-6}}{2!}=\frac{0.0892}{2}  =0.04462\\\\\\P(x\geq3)=1-(0.00248+0.01487+0.04462)=1-0.06197=0.938

    c) 2<x<7

    P(2 < X < 7)= P(x=3)+P(x=4)+P(x=5)+P(x=6)\\\\P(x=3)=0.0892\\\\P(x=4)=0.1339\\\\P(x=5)=0.1606\\\\P(x=6)=0.1606\\\\\\P(2 < X < 7)=0.0892+0.1339+0.1606+0.1606=0.5443

    d) and e)

    \mu=\lambda=6\\\\\\\sigma=\sqrt{\lambda}=2.45

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45:7+7-4:2-5:5*4+35:2 =? ( )