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A catalog company that receives the majority of its orders by telephone conducted a study to...

Question:

A catalog company that receives the majority of its orders by telephone conducted a study to determine how long customers were willing to wait on hold before ordering a product. The length of time was found to be a random variable best approximated by an exponential distribution with a mean equal to 2.8 minutes. What proportion of callers are put on hold longer than 2.8 minutes?

a. 0.6321

b. 0.3678

c. 0.6081

d. 0.50

Exponential distribution:


The exponential distribution is also known by the name of memory less distribution because it is the only continuous distribution that has the property of memory less. This distribution has only one parameter\lambda . The cumulative distribution function {eq}F\left( {X;\lambda } \right) {/eq} of this distribution is {eq}1 - {e^{ - \lambda \left( x \right)}} {/eq}.

Answer and Explanation:


The random variable X represents the waiting time of callers who put on hold before ordering a product that follows exponential distribution with {eq}\lambda {/eq}.

The customer's average waiting time {eq}\left( \beta \right) {/eq} is 2.8 minutes.

The parameter {eq}\lambda {/eq} is:

{eq}\begin{align*} \lambda &= \dfrac{1}{\beta }\\ &= \dfrac{1}{{2.8}}{\rm{minutes}} \end{align*} {/eq}


Calculate the proportion of those callers that are put on hold longer than 2.8 minutes:

{eq}\begin{align*} P\left( {X > 2.8} \right) &= 1 - P\left( {X \le 2.8} \right)\\ &= 1 - \left( {1 - {e^{ - 2.8\left( {\dfrac{1}{{2.8}}} \right)}}} \right)\\ &= 1 - \left( {1 - {e^{ - \dfrac{{2.8}}{{2.8}}}}} \right)\\ &= 1 - \left( {1 - {e^{ - 1}}} \right)\\ &= 1 - 1 + {e^{ - 1}}\\ &= {e^{ - 1}}\\ &= 0.3678 \end{align*} {/eq}

Thus, the proportion of callers that are put on hold longer than 2.8 minutes is 0.3678

Hence, the correct option is b.


Learn more about this topic:

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Exponential Growth: Definition & Examples

from High School Algebra I: Help and Review

Chapter 6 / Lesson 10
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