Respuesta :
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:
[tex]f(x)=0.000121e^{-0.000121x};\ x>0[/tex]
Compute the value of P (X < 5730) as follows:
[tex]\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[/tex]
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%.
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([tex]\lambda[/tex]) where, [tex]\lambda[/tex] = 0.000121
The probability distribution function of exponential distribution is;
f(x) = [tex]\lambda e^{-\lambda x}[/tex] where, x > 0
Similarly, CDF of exponential distribution is;
P(X <= x) = 1 - [tex]e^{-\lambda x}[/tex] where, x > 0
So, P(X < 5,730) = 1 - [tex]e^{-0.000121* 5730}[/tex] = 1 - 0.499 = 0.5
Therefore, probability of amount (percent of one gram) of Carbon-14 lasting less than 5,730 years is 50% .
