Question

Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find P(x < 5,730).

a. Sketch the graph, and shade the area of interest

b. Find the probability. P(x < 5,730) = __________

Short answer

a. The amount of carbon-14 is$0.000121$

b. The answer of probability$P(x<5,730)$= $0.50009$

Step-by-step solution

Explanation the  amount of carbon-14  trough  graph

It has been provided that $X~Exp(0.000121)$

So the mean value is $0.000121$

The formula of probability density is $f\left(x\right)=m{e}^{-mx}$

Applying all data in the above equation

$f\left(x\right)=0.000121e^{0.000121x}$

At $x=0$, the $f\left(x\right)$gain maximum value and lie in y axis. then $f\left(x\right)\to f\left(0\right)$

$\begin{array}{rcl}f\left(0\right)& =& 0.000121e^{0.000121\times 0}\\ & =& 0.000121\end{array}$

This value is equal to the mean value. So the maximum value is equal mean value (m) and the point is provided in the graph.

Step  2:  Formula of cumulative distribution 

In the exponential distribution, the cumulative distribution is $P(X<x)=1-e^{-mx}$

here m = 0.000121

Applying all data in the above equation

$\begin{array}{rcl}P(x& <& 5730)=1-e^{0.000121\times 5730}\\ & =& 1-e^{0.699333}\\ & =& 1-0.49991\\ & =& 0.50009\end{array}$