Question

Based on data collected from metal shredders across the nation, the amount L of extractable lead in metal shredder residue has an approximate exponential distribution with mean μ = 2.0 milligrams per liter. What is the probability that L is greater than 1.20 milligrams per liter?

Answer 1

Answer:

$P(L >1.2)$

The cumulative distirbution function is given by:

$P(L <l) = 1-e^{-\lambda x}$

And using the complement rule we have:

$P(L>1.2) = 1-P(L<1.2) = 1-[1- e^{-\frac{1.2}{2}}] = e^{-\frac{1.2}{2}} = 0.549$

Step-by-step explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution". The probability density function is given by:

$P(X)=\lambda e^{-\lambda x}$

Solution to the problem

For this case we define the random variable L="the amount L of extractable lead in metal shredder residue" and the distirbution for X is given by:

$L \sim Exp( \lambda =\frac{1}{2}=0.5$

For this case we want this probability:

$P(L >1.2)$

The cumulative distirbution function is given by:

$P(L <l) = 1-e^{-\lambda x}$

And using the complement rule we have:

$P(L>1.2) = 1-P(L<1.2) = 1-[1- e^{-\frac{1.2}{2}}] = e^{-\frac{1.2}{2}} = 0.549$

Answer 2

Answer:

Probability that L is greater than 1.20 milligrams per liter is 0.5488.

Step-by-step explanation:

We are given that based on data collected from metal shredders across the nation, the amount L of extractable lead in metal shredder residue has an approximate exponential distribution with mean μ = 2.0 milligrams per liter.

The probability distribution for exponential distribution is given by;

                      $f(l) = \lambda e^{-\lambda l} ; l >0$

where, $\lambda$ = parameter of this distribution

Let L = Amount of extractable lead in metal shredder residue

Now, as we know that the mean of exponential distribution is;

    Mean, $\mu$ = $\frac{1}{\lambda}$   ⇒ 2.0 = $\frac{1}{\lambda}$     {because we are given with the mean}

So, $\lambda=\frac{1}{2}$ = 0.5

Hence, L ~ Exp($\lambda = 0.5$)

Now, to find the less than or greater than probabilities in exponential distribution we use the Cumulative distribution function of exponential function, i.e.;

             $F(L) = P(L \leq l) = 1 - e^{-\lambda l} ; l >0$

So, probability that L is greater than 1.20 milligrams per liter is given by = P(L > 1.20 milligrams per liter)

      P(L > 1.20) = 1 - P(L $\leq$ 1.20)

                        = 1 - [ $1 - e^{-0.5 \times 1.20}$ ]

                        = $e^{-0.5 \times 1.20}$ = 0.5488

Therefore, probability that L is greater than 1.20 milligrams per liter is 0.5488.