Question

Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of Carbon-14. We are interested in the time (years) it takes to decay Carbon-14. Find the amount (percent of one gram) of Carbon-14 lasting less than 5,730 years. This means, find the probability P(x < 5,730).

Answer 1

Answer:

P(x < 5,730) = 0.5

Step-by-step explanation:

We are given that Carbon-14 is said to decay exponentially. The decay rate is 0.000121.

Let X = amount (percent of one gram) of Carbon-14

So, X ~ Exp($\lambda$)  where, $\lambda$ = 0.000121

The probability distribution function of exponential distribution is;

      f(x) = $\lambda e^{-\lambda x}$  where, x > 0

Similarly, CDF of exponential distribution is;

  P(X <= x) = 1 - $e^{-\lambda x}$ where, x > 0

So, P(X < 5,730) = 1 - $e^{-0.000121* 5730}$ = 1 - 0.499 = 0.5

Therefore, probability of amount (percent of one gram) of Carbon-14 lasting less than 5,730 years is 50% .

 

Answer 2

Answer:

The percentage of Carbon-14 lasting less than 5,730 years is 50%.

Step-by-step explanation:

Let X = number of years it takes for Carbon-14 to decay.

The random variable X follows an Exponential distribution with decay parameter, λ = 0.000121.

The probability mass function of X is:

$f(x)=0.000121e^{-0.000121x};\ x>0$

Compute the value of P (X < 5730) as follows:

$\int\limits^{5730}_{0} {f(x)} \, dx =\int\limits^{5730}_{0} {0.000121e^{-0.000121x}} \, dx \\=0.000121\int\limits^{5730}_{0} {e^{-0.000121x}} \, dx \\=0.000121 |\frac{e^{-0.000121x}}{-0.000121}| ^{5730}_{0}\\=[-e^{-0.000121\times5730}+e^{-0.000121\times0}]\\=1-0.4999\\=0.5000$

The percentage of Carbon-14 lasting less than 5,730 years is,

0.5000 × 100 = 50%

Thus, the percentage of Carbon-14 lasting less than 5,730 years is 50%.