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Statistics / Statistics for Engineers and Scientists 4 / Chapter 4 / Problem 23SE

Statistics for Engineers and Scientists | 4th Edition | ISBN: 9780073401331 | Authors: William Navidi

A distribution that has been used to model tolerance

Chapter 4, Problem 23SE

(choose chapter or problem)

QUESTION:

A distribution that has been used to model tolerance levels in bioassays is the logistic distribution with parameters $\alpha{}$ and $\beta{}.$ The cumulative distribution function of the logistic distribution is

$F(x)=\left[1+e^{-(x-\alpha / \beta}\right]^{-1}$

The parameter $\alpha{}$ may be any real number; the parameter $\beta{}$ may be any positive number. Let $X$ be a random variable with this distribution.

a. Find the probability density function $f_{x}(x)$.

b. Show that $f_{x}(x)$ is symmetric around α, that is, $f_{x}(\alpha-x)=f x(\alpha+x)$ for all x.

c. Explain why the symmetry described in part (b) shows that $\mu_{x}=\alpha$. You may assume that $\mu_{x}$ exists.

Equation transcription:

$F(x)=[1+{e}^{-(x-\alpha{}/\beta{}}{]}^{-1}$

${f}_{x}(x)$

${f}_{x}(\alpha{}-x)=fx(\alpha{}+x)$

${\mu{}}_{x}=\alpha{}$

${\mu{}}_{x}$

Text transcription:

F(x)=\left[1+e^{-(x-\alpha / \beta}\right]^{-1}

f_{x}(x)

f_{x}(\alpha-x)=f x(\alpha+x)

\mu_{x}=\alpha

\mu_{x}

Questions & Answers

QUESTION:

$F(x)=\left[1+e^{-(x-\alpha / \beta}\right]^{-1}$

The parameter $\alpha{}$ may be any real number; the parameter $\beta{}$ may be any positive number. Let $X$ be a random variable with this distribution.

a. Find the probability density function $f_{x}(x)$.

b. Show that $f_{x}(x)$ is symmetric around α, that is, $f_{x}(\alpha-x)=f x(\alpha+x)$ for all x.

c. Explain why the symmetry described in part (b) shows that $\mu_{x}=\alpha$. You may assume that $\mu_{x}$ exists.

Equation transcription:

$F(x)=[1+{e}^{-(x-\alpha{}/\beta{}}{]}^{-1}$

${f}_{x}(x)$

${f}_{x}(\alpha{}-x)=fx(\alpha{}+x)$

${\mu{}}_{x}=\alpha{}$

${\mu{}}_{x}$

Text transcription:

F(x)=\left[1+e^{-(x-\alpha / \beta}\right]^{-1}

f_{x}(x)

f_{x}(\alpha-x)=f x(\alpha+x)

\mu_{x}=\alpha

\mu_{x}

ANSWER:

Answer:

Step 1 of 3:

(a)

In this question, we are asked to find the probability density function $f(x)$ for given cumulative distribution function of logistic distribution.

$F(x)\ =\ [1\ +\ {e}^{-(x\ -\ \alpha{})/\beta{}}{]}^{-1}$

Where $\alpha{}$ may be any real number; the parameter $\beta{}$ may be any positive number..

Let $X$ be a random variable with this distribution.

We know the formula for calculating the cumulative distribution function (CDF) is:

$F(x)\ =\ \int\limits_{-\infty{}}^{x}f(t)dt$

Then probability density function can be found as a $f(x)$ = $\frac{dF(x)}{dx}$

$f(x)$ = $\frac{dF(x)}{dx}$

= $\frac{d}{dx}($ $\ [1\ +\ {e}^{-(x\ -\ \alpha{})/\beta{}}{]}^{-1}$ ) assume t = $\ [1\ +\ {e}^{-(x\ -\ \alpha{})/\beta{}}$ ]

= $\frac{d}{dt}($ $t{)}^{-1}$ $\times{}$ $\frac{d}{dx}$ $\ [1\ +\ {e}^{-(x\ -\ \alpha{})/\beta{}}$ ] (using chain rule of differentiation)

= $\frac{-1}{{t}^{2}}$ $\times{}$ [ $\frac{d}{dx}$ $\ (1)\ +\frac{d}{dx}\ {e}^{-(x\ -\ \alpha{})/\beta{}}$ ]

= $\frac{-1}{{t}^{2}}$ $\times{}$ [ $0-\frac{{e}^{-(x\ -\ \alpha{})/\beta{}}}{\beta{}}$ ]

= $\frac{{e}^{-(x\ -\ \alpha{})/\beta{}}}{\beta{}\times{}\ [1\ +\ {e}^{-(x\ -\ \alpha{})/\beta{}}{]}^{2}}$ ……………(1)

Hence probability distribution function $f(x)$ = $\frac{{e}^{-(x\ -\ \alpha{})/\beta{}}}{\beta{}\times{}\ [1\ +\ {e}^{-(x\ -\ \alpha{})/\beta{}}{]}^{2}}$

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