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

Answer:

(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

Step-by-step 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:

$P(X=x)=\frac{e^{-\lambda}\lambda^{x}}{x!};\ x=0, 1, 2,...$

(a)

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

$P(X=7)=\frac{e^{-6}(6)^{7}}{7!}=0.1377$

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)

              $=1-\frac{e^{-6}(6)^{0}}{0!}-\frac{e^{-6}(6)^{1}}{1!}-\frac{e^{-6}(6)^{2}}{2!}\\=1-0.00248-0.01487-0.04462\\=0.93803\\\approx0.9380$

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)

                    $=\frac{e^{-6}(6)^{3}}{3!}+\frac{e^{-6}(6)^{4}}{4!}+\frac{e^{-6}(6)^{5}}{5!}+\frac{e^{-6}(6)^{6}}{6!}\\=0.08924+0.13385+0.16062+0.16062\\=0.54433\\\approx0.5443$

Thus, the value of P (2 < X < 7) is 0.5443.

(d)

The standard deviation of a Poisson distribution is:

$SD(X)=\sigma_{x}=\sqrt{\lambda}$

Compute the value of standard deviation of X as follows:

$\sigma_{x}=\sqrt{6}=2.4495$

Thus, the standard deviation of X is 2.4495