Chapter 2.4, Problem 22E

a.

To determine

Find the probability that the concentration is greater than 0.5.

Answer to Problem 22E

The probability of the concentration is greater than 0.5 is 0.269.

Explanation of Solution

Given info:

The concentration of a reactant is a random variable with probability density function

$f\left(x\right)=\{\begin{array}{l}\frac{2e^{-2x}}{1-e^{-2}},\text{ if }0<x<\text{1}\\ 0,\text{ Otherwise}\end{array}$

Calculation:

Let X be a continuous random variable with probability density function $f\left(x\right)$. Let b be a number and $X>a$ then the probability of $X>a$ can be obtained as,

$P\left(X\ge a\right)=P\left(X>a\right)={\displaystyle {\int }_a^{\infty }f\left(x\right)dx}$

Let X is the concentration of a reactant.

The probability that the concentration is greater than 0.5can be obtained as,

$\begin{array}{c}P\left(X>0.5\right)={\displaystyle {\int }_{0.5}^1\frac{2e^{-2x}}{1-e^{-2}}}dx\\ =2{\left[-\frac{e^{-2x}}{2\left(1-e^{-2}\right)}\right]}_{0.5}^1\\ =-\frac{e^{-2\left(1\right)}}{\left(1-e^{-2}\right)}+\frac{e^{-2\left(0.5\right)}}{\left(1-e^{-2}\right)}\\ =-0.1565+0.4255\end{array}$

$=0.269$

Thus, the probability of the concentration is greater than 0.5 is 0.269.

b.

To determine

Find the mean concentration.

Answer to Problem 22E

The mean concentration is 0.3435.

Explanation of Solution

Calculation:

Mean:

Let X be a continuous random variable with probability density function $f\left(x\right)$ then the mean of X is,

${\mu }_X={\displaystyle {\int }_{-\infty }^{\infty }xf\left(x\right)dx}$

The mean concentration can be obtained as,

$\begin{array}{c}{\mu }_X={\displaystyle {\int }_0^{1}x\left(\frac{2e^{-2x}}{1-e^{-2}}\right)}dx\\ =2\left(\frac{e^2}{e^2-1}\right){\displaystyle {\int }_0^{1}x{e}^{-2x}}dx\end{array}$

$\begin{array}{c}\text{Let }2x=u\\ 2dx=du\\ dx=\frac{1}{2}du\end{array}$

Upper limit: if $x=0\Rightarrow \frac{u}{2}=0\Rightarrow u=0$

Lower limit: if $x=1\Rightarrow \frac{u}{2}=1\Rightarrow u=2$

$\begin{array}{c}{\mu }_X=\frac{1}{2}\left(\frac{e^2}{e^2-1}\right){\displaystyle {\int }_0^{2}u{e}^{-u}}du\\ =\frac{1}{2}\left(\frac{e^2}{e^2-1}\right)\left({\left[-u{e}^{-u}\right]}_0^2+{\displaystyle {\int }_0^2{e}^{-u}du}\right)\\ =\frac{1}{2}\left(\frac{e^2}{e^2-1}\right)\left(\left(-2e^{-2}-0\right)+{\left[-e^{-u}\right]}_0^2\right)\\ =\frac{1}{2}\left(\frac{e^2}{e^2-1}\right)\left(\left(-2e^{-2}-0\right)+\left[-e^{-2}+1\right]\right)\end{array}$

      $\begin{array}{c}=\frac{1}{2}\left(\frac{e^2}{e^2-1}\right)\left(-3e^{-2}+1\right)\\ =\frac{1}{2}\left(\frac{e^2-3}{e^2-1}\right)\\ =0.3435\end{array}$

Thus, the mean concentration is 0.3435.

c.

To determine

Find the probability that the concentration is within $\pm 0.1$ of the mean.

Answer to Problem 22E

The probability that the concentration is within $\pm 0.1$ of the mean is 0.2342.

Explanation of Solution

Calculation:

The probability that the concentration is within $\pm 0.1$ of the mean can be obtained as,

$\begin{array}{c}P\left(\mu -0.1<X<\mu +0.1\right)={\displaystyle {\int }_{0.2435}^{0.4435}\frac{2e^{-2x}}{1-e^{-2}}}dx\\ =2{\left[-\frac{e^{-2x}}{2\left(1-e^{-2}\right)}\right]}_{0.2435}^{0.4435}\\ =-\frac{e^{-2\left(0.4435\right)}}{\left(1-e^{-2}\right)}+\frac{e^{-2\left(0.2435\right)}}{\left(1-e^{-2}\right)}\\ =-0.4764+0.7106\end{array}$

                                    $=0.2342$

Thus, the probability that the concentration is within $\pm 0.1$ of the mean is 0.2342.

d.

To determine

Find the standard deviation of the concentration.

Answer to Problem 22E

The standard deviation of the concentration is 0.2627.

Explanation of Solution

Calculation:

Variance:

Let X be a continuous random variable with probability density function $f\left(x\right)$ then the variance of X is,

${\sigma }_{{}_X}^2={\displaystyle {\int }_{-\infty }^{\infty }{x}^{2}f\left(x\right)dx-{\mu }_X^2}$

Standard deviation:

The standard deviation is of the continuous variable is,

${\sigma }_X=\sqrt{{\sigma }_X^2}$

The variance of the concentration can be obtained as,

$\begin{array}{c}{\sigma }_{{}_X}^2={\displaystyle {\int }_0^1{x}^2\left(\frac{2e^{-2x}}{1-e^{-2}}\right)}dx-{\mu }^2\\ =\left(\frac{e^2}{e^2-1}\right){\displaystyle {\int }_0^{1}2x^2{e}^{-2x}}dx-{\left(0.3435\right)}^2\end{array}$

$\begin{array}{c}\text{Let }2x=u\\ x^2=\frac{u^2}{4}\\ 2dx=du\\ dx=\frac{1}{2}du\end{array}$

Upper limit: if $x=0\Rightarrow \frac{u}{2}=0\Rightarrow u=0$

Lower limit: if $x=1\Rightarrow \frac{u}{2}=1\Rightarrow u=2$

$\begin{array}{c}{\sigma }_{{}_X}^2=\frac{1}{4}\left(\frac{e^2}{e^2-1}\right){\displaystyle {\int }_0^2{u}^2{e}^{-u}}du-0.1180\\ =\frac{1}{4}\left(\frac{e^2}{e^2-1}\right)\left({\left[-u^2{e}^{-u}\right]}_0^2+{\displaystyle {\int }_0^{2}2u{e}^{-u}du}\right)-0.1180\\ =\frac{1}{4}\left(\frac{e^2}{e^2-1}\right)\left(\left(-4e^{-2}-0\right)+2\left({\left[-u{e}^{-u}\right]}_0^2+{\displaystyle {\int }_0^2{e}^{-u}du}\right)\right)-0.1180\\ =\frac{1}{4}\left(\frac{e^2}{e^2-1}\right)\left(\left(-4e^{-2}-0\right)+2\left(-2e^{-2}-0\right)+2\left[-e^{-2}+1\right]\right)-0.1180\end{array}$

       $\begin{array}{c}=\frac{1}{4}\left(\frac{e^2}{e^2-1}\right)\left(-4e^{-2}-4e^{-2}-2e^{-2}+2\right)-0.1180\\ =\frac{1}{4}\left(\frac{e^2}{e^2-1}\right)\left(-10e^{-2}+2\right)-0.1180\\ =\frac{1}{2}\left(\frac{e^2-5}{e^2-1}\right)-0.1180\\ =0.1870-0.1180\end{array}$

$=0.069$

Thus, the variance of the concentration is 0.04.

The standard deviation of the concentration is,

$\begin{array}{c}{\sigma }_X=\sqrt{0.069}\\ =0.2627\end{array}$

Thus, the standard deviation of the concentration is 0.2627.

d.

To determine

Find the probability that the concentration is within $\pm \sigma$ of the mean.

Answer to Problem 22E

The probability that the concentration is within $\pm \sigma$ of the mean is 0.6399.

Explanation of Solution

Calculation:

The probability that the concentration is within $\pm \sigma$ of the mean can be obtained as,

$\begin{array}{c}P\left(\mu -\sigma <X<\mu +\sigma \right)=P\left(0.3435-0.2627<X<0.3435+0.2627\right)\\ =P\left(0.0808<X<0.6062\right)\\ ={\displaystyle {\int }_{0.0808}^{0.6062}\frac{2e^{-2x}}{1-e^{-2}}}dx\\ =2{\left[-\frac{e^{-2x}}{2\left(1-e^{-2}\right)}\right]}_{0.0808}^{0.6062}\\ =-\frac{e^{-2\left(0.6062\right)}}{\left(1-e^{-2}\right)}+\frac{e^{-2\left(0.0808\right)}}{\left(1-e^{-2}\right)}\end{array}$

$\begin{array}{c}=-0.3440+0.9839\\ =0.6399\end{array}$

Thus, the probability that the concentration is within $\pm \sigma$ of the mean is 0.6399.

f.

To determine

Find the cumulative distribution function of the concentration.

Answer to Problem 22E

The cumulative distribution function of the concentration is,

$F\left(x\right)=\{\begin{array}{l}0,\text{ if }x\le \text{0}\\ \frac{1-e^{-2x}}{1-e^{-2}}\text{, if }0<x<1\\ 1,\text{ if }x\ge 1\end{array}$

Explanation of Solution

Calculation:

Cumulative distribution function:

Let X be a continuous random variable with probability density function $f\left(x\right)$ then the cumulative distribution function of X is,

$F\left(x\right)=P\left(X\le x\right)={\displaystyle {\int }_{-\infty }^{x}f\left(t\right)dt}$

For $x\le \text{0}$:

$\begin{array}{c}F\left(x\right)={\displaystyle {\int }_{-\infty }^{x}0}dt\\ =0\end{array}$

For $0<x<1$:

$\begin{array}{c}F\left(x\right)={\displaystyle {\int }_0^x\frac{2e^{-2t}}{1-e^{-2}}}dt\\ =2{\left[-\frac{e^{-2t}}{2\left(1-e^{-2}\right)}\right]}_0^x\\ =-\left(\frac{e^{-2\left(x\right)}-e^{-2\left(0\right)}}{1-e^{-2}}\right)\\ =\frac{1-e^{-2x}}{1-e^{-2}}\end{array}$

For $x\ge 1$:

$\begin{array}{c}F\left(x\right)={\displaystyle {\int }_0^1\frac{2e^{-2t}}{1-e^{-2}}}dt\\ =2{\left[-\frac{e^{-2t}}{2\left(1-e^{-2}\right)}\right]}_0^1\\ =-\left(\frac{e^{-2\left(1\right)}-e^{-2\left(0\right)}}{1-e^{-2}}\right)\\ =\frac{1-e^{-2}}{1-e^{-2}}\end{array}$

$=1$

Thus, the cumulative distribution function of the concentration is,

$F\left(x\right)=\{\begin{array}{l}0,\text{ if }x\le \text{0}\\ \frac{1-e^{-2x}}{1-e^{-2}}\text{, if }0<x<1\\ 1,\text{ if }x\ge 1\end{array}$