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 that 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.Compute P(X = 7). Round your answer to four decimal places.Compute P(X ≥ 3). Round your answer to four decimal places.Compute P(2 < X < 7). Round your answer to four decimal places.Compute σX, the standard deviation of X. Round your answer to four decimal places.

Answer 1

(a) The value of P (X = 7) is 0.1377.

(b) The value of P (X ≥ 3) is 0.9380.

(c) The value of P (2 < X < 7) is 0.5443.

(d) The standard deviation of X is 2.4495

Explanation:

Let X = the number of tracks counted in 1 cm² of surface area.

It is provided that the average number of tracks is 6 per cm².

The random variable
`X\sim Poisson(\lambda=6)`.

The probability function of a Poisson distribution is:

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

(a)

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

`P(X=7)=(e^(-6)(6)^(7))/(7!)=0.1377`

Thus, the value of P (X = 7) is 0.1377.

(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 0.5443.

(d)

The standard deviation of a Poisson distribution is:

`SD(X)=\sigma_(x)=√(\lambda)`

Compute the value of standard deviation of X as follows:

`\sigma_(x)=√(6)=2.4495`

Thus, the standard deviation of X is 2.4495