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Molecules of a toxic chemical eventually decompose into inert substances. Suppose the decomposition time is exponentially distributed with a mean of 1/lambda. The half-life of such a persistent poison is that time beyond which the probability is .50 that a particular molecule will remain toxic. Find the half-life for chemicals whose molecules have an average decomposition time of (a) 6 yearshalflife= ? years(b) 40 yearshalflife= years(c) 344 yearshalflife= ? years(d) 5584 yearshalflife= ? years

Answer :

rodolforodriguezr

Answer:

a) 4.16 years

b) 27.73 years

c) 238.44 years

d) 3,870.53 years

Step-by-step explanation:

Let X be the random variable that measures the decomposition time.

a)

[tex] \bf \lambda =6[/tex]

In this case, since the decomposition time is exponentially distributed with a mean of 1/6, we have

[tex] \bf P(X\leq t)=1-e^{-t/6}\Rightarrow P(X>t)=1-(1-e^{-t/6})=e^{-t/6}[/tex]

and we must find a t such that P(x>t)=0.5.

[tex] \bf P(X>t)=0.5\Rightarrow e^{-t/6}=0.5\Rightarrow -t/6=ln(0.5)\Rightarrow t=-6ln(0.5)=4.16\;years[/tex]

b)

[tex] \bf \lambda =40[/tex]

[tex] \bf P(X>t)=0.5\Rightarrow e^{-t/40}=0.5\Rightarrow -t/40=ln(0.5)\Rightarrow t=-40ln(0.5)=27.73\;years[/tex]

c)

[tex] \bf \lambda =344[/tex]

[tex] \bf P(X>t)=0.5\Rightarrow e^{-t/344}=0.5\Rightarrow -t/344=ln(0.5)\Rightarrow t=-344ln(0.5)=238.44\;years[/tex]

d)

[tex] \bf \lambda =5584[/tex]

[tex] \bf P(X>t)=0.5\Rightarrow e^{-t/5584}=0.5\Rightarrow -t/5584=ln(0.5)\Rightarrow t=-5584ln(0.5)=3870.53\;years[/tex]

MrRoyal

The half lives for the decomposition times are 4.16, 27.73, 238.44 and 3870.53 years, respectively

How to determine the half lives?

The molecules of the toxic chemicals follow an exponential distribution.

Such that

p = 0.50 and Mean = 1/λ

An exponential distribution is represented as:

[tex]P(x \le t) = 1 - e^{-\lambda t}[/tex]

Using the complement rule, we have:

[tex]P(x > t) = 1 - P(x \le t)[/tex]

Substitute known values

[tex]P(x > t) = 1 - 1 + e^{-\lambda t}[/tex]

Evaluate the difference

[tex]P(x > t) = e^{-\lambda t}[/tex]

Substitute 0.5 for P(x > t)

[tex]e^{-\lambda t} = 0.5[/tex]

Take the natural logarithm of both sides

[tex]-\lambda t = \ln(0.5)[/tex]

Make t the subject

[tex]t = -\frac{1}{\lambda} * \ln(0.5)[/tex]

When average decomposition time = 6 years, the half life is:

[tex]t = -6 * \ln(0.5)[/tex]

[tex]t = 4.16[/tex]

When average decomposition time = 40 years, the half life is:

[tex]t = -40 * \ln(0.5)[/tex]

[tex]t = 27.73[/tex]

When average decomposition time = 344 years, the half life is:

[tex]t = -344 * \ln(0.5)[/tex]

[tex]t = 238.44[/tex]

When average decomposition time = 5584 years, the half life is:

[tex]t = -5584 * \ln(0.5)[/tex]

[tex]t = 3870.53[/tex]

Hence, the half lives for the decomposition times are 4.16, 27.73, 238.44 and 3870.53 years, respectively

Read more about exponential distribution at:

https://brainly.com/question/14904292

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