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 cm2 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 cm2 of surface area.

a)Find P(X = 7).
b)Find P(X ≥ 3).
c)Find P(2 < X < 7).
d)Find μX.
e)Find σX

Answer 1

(a) The value of P (X = 7) is 0.1388.

(b) The value of P (X ≥ 3) is 0.9380.

(c) The value of P (2 < X < 7) is 0.5433.

(d)
`\mu_(X)=6`

(e)
`\sigma_(X)=2.45`

Explanation:

Let X = number of uranium fission tracks on per cm² surface area of the mineral.

The average number of track per cm² surface area is, λ = 6.

The random variable X follows a Poisson distribution with parameter λ = 6.

The probability mass function of a Poisson distribution is:

`P(X=x)=(e^(-\lambda)\lambda^(x))/(x!);\ x=0, 1, 2, 3...`

(a)

Compute the value of P (X = 7) as follows:

`P(X=6)=(e^(-6)(6)^(7))/(7!)=(0.0025* 279936)/(5040)=0.1388`

Thus, the value of P (X = 7) is 0.1388.

(b)

Compute the value of P (X ≥ 3) as follows:

P (X ≥ 3) = 1 - P (X < 3)

= 1 - P (X = 0) - P (X = 1) - P (X = 2)

`=1-(e^(-6)(6)^(0))/(0!)-(e^(-6)(6)^(1))/(1!)-(e^(-6)(6)^(2))/(2!)\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380`

Thus, the value of P (X ≥ 3) is 0.9380.

(c)

Compute the value of P (2 < X < 7) as follows:

P (2 < X < 7) = P (X = 3) + P (X = 4) + P (X = 5) + P (X = 6)

`=(e^(-6)(6)^(3))/(3!)+(e^(-6)(6)^(4))/(4!)+(e^(-6)(6)^(5))/(5!)+(e^(-6)(6)^(6))/(6!)\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443`

Thus, the value of P (2 < X < 7) is 0.5433.

(d)

The mean of the Poisson distribution is:

`\mu_(X)=\lambda=6`

(e)

The standard deviation of the Poisson distribution is:

`\sigma_(X)=\sqrt{\sigma^(2)_(X)}=√(\lambda)=√(6)=2.4495\approx2.45`