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Chapter 4: Q102SE (page 194)

Let X represent the number of individuals who respond to a particular online coupon offer. Suppose that X has approximately a Weibull distribution with \(\alpha = 10\)and \(\beta = 20\). Calculate the best possible approximation to the probability that X is between \(15 and 20\), inclusive.

Short Answer

Expert verified

\(P(15 \le X \le 20) = {e^{ - 59049/1048576}} - \frac{1}{e} \approx 0.57736 = 57.736\% \)

Step by step solution

01

Definition of Outcomes

An outcome is a possible consequence of experimentation or trial in probability theory. Each conceivable experiment outcome is distinct, and alternative outcomes are necessarily exclusive (only one outcome will occur on each trial of the experiment).

02

Calculation of best possible approximation to the probability.

Given: X has a Weiull distribution with

\(\begin{array}{l}\alpha = 10\\\beta = 20\\P(15 \le X \le 20)\end{array}\)

The cumulative distribution function of a Weibull distribution is:

\(F(x;\alpha ,\beta ) = \left\{ {\begin{array}{*{20}{c}}0&{x < 0}\\{1 - {e^{ - {{(x/\beta )}^\alpha }}}}&{x \ge 0}\end{array}} \right.\)

The probability between two boundaries is the difference of the cumulative distribution function evaluated at the boundaries:

\(\begin{array}{l}P(15 \le X \le 20) = F(20) - F(15)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = F(20;10,20) - F(15;10,20)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 - {e^{ - {{(20/20)}^{10}}}} - \left( {1 - {e^{ - {{(15/20)}^{10}}}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {e^{ - 59049/1048576}} - \frac{1}{e} \approx 0.57736 = 57.736\% \end{array}\)

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