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

Answer:

The required answer is shown below:

Step-by-step explanation:

Consider the provided information.

The formula for calculating poisson probability mass function: $P(k)=\frac{\lambda^ke^{-\lambda} }{k!}$

Where λ is average number of events, the value of e = 2.718..

K can take the values 0,1,2...

Part (a)Find P(X = 7).

A certain mineral specimen is of such an age that there should be an average of 6 tracks per cm² of surface area.

Use the above formula and substitute x=7 and λ=6

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

$P(X=7)=0.1377$

Hence, P(X = 7) = 0.1377

Part (b)Find P(X ≥ 3).

This can be calculated as:

$P(X\geq 3)=1-P(X=0)-P(X=1)-P(X=2)$

$P(X\geq 3)=1-\frac{6^0e^{-6} }{0!}-\frac{6^1e^{-6}}{1!}-\frac{6^2e^{-6}}{2!}$

$P(X=7)=0.9380$

Hence, P(X ≥ 3) = 0.9380

Part (c)Find P(2 < X < 7).

The sum can be calculated as:

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

$P(2<X<7)=\frac{6^3e^{-6} }{3!}+\frac{6^4e^{-6}}{4!}+\frac{6^5e^{-6}}{5!}+\frac{6^6e^{-6}}{6!}$

$P(2<X<7)=0.5443$

Hence, P(2 < X < 7)= 0.5443

Part (d)Find μX.

If λ is average number of successes or region in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to λ.

$\mu_x=\lambda=6$

Part (e)Find σX

If λ is average number of successes or region in the Poisson distribution, then the mean and the variance of the Poisson distribution are both equal to λ.

$(\sigma_x)^2=\lambda$

$\sigma_x=\sqrt{\lambda}$

$\sigma_x=\sqrt{6}=2.449$