Question

Radioactive isotopes are commonly used to estimate artifact ages. Let X be the time in years for a Carbon-14 atom to decay to Nitrogen-14. X can be modeled as an exponential random variable with λ=1/8267
a. What are the units of What are the units of λ?
b. What is the probability the decay time is between 5000 and 6000 years?
c. Compute SD(X).
d. Compute the median decay time. The result is called the half-life of Carbon-14?

Answer 1

Answer:

Step-by-step explanation: Attached Picture.

Answer 2

Answer:

a) 1 / years

b) P ( 5000 < X < 6000) =  0.06222

c) 8267 years

d) T_1/2 = 5730.25 years          

Step-by-step explanation:

Given:

- The decay constant λ = 1 / 8267

- X is an exponential random Variable

Find:

a. What are the units of What are the units of λ?

b. What is the probability the decay time is between 5000 and 6000 years?

c. Compute SD(X).

d. Compute the median decay time. The result is called the half-life of Carbon-14?

Solution:

- The decay constant λ means the rate at which the C-14 atoms decays into N-14 atom. The rate is expressed in units of 1 / year.

- The random variable X follows an exponential distribution which has a probability mass function and cumulative density functions as follows:

                    P ( X = t ) = λ*e^(-λ*t)

                    P ( X = t ) = e^(-t/8267) / 8267

                    P ( X < t ) = 1 - e^(-λ*t) = 1 - e^(-t/8267)

- The probability of the decay time between years 5000 and 6000 years is:

                    P ( 5000 < X < 6000) = 1 - e^(-6000/8267) - 1 + e^(-5000/8267)

                    P ( 5000 < X < 6000) =  0.06222

- The standard deviation of the exponential distribution is given by:

                    SD(X) = 1 / λ = 8267 years

- The median decay time for an exponential distribution is given by:

                    T_1/2 = Ln(2) / λ = Ln(2)*8267

                    T_1/2 = 5730.25 years