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

Answer 1

a) 0.138

b) 0.938

c) 0.544

d) 6

e) 2.45

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)=(\lambda^ke^(-\lambda))/(k!)`

Then

a) x=7

`P(x=7)=(\lambda^7e^(-\lambda))/(7!)=(6^7e^(-6))/(7!)=(693.89)/(5040)=0.138`

b) x≥3

`P(x\geq 3)=1-(P(0)+P(1)+P(2))\\\\P(0)=(6^0e^(-6))/(0!)=(0.00248)/(1) =0.00248\\\\P(1)=(6^1e^(-6))/(1!)=(0.01487)/(1) =0.01487\\\\P(2)=(6^2e^(-6))/(2!)=(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=√(\lambda)=2.45`