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A 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 waiting time was found to be a variable best approximated by an exponential distribution with a mean length of waiting time equal to 3 minutes (i.e. the mean number of calls answered in a minute is 1/3). What proportion of customers having to hold more than 4.5 minutes will hang up before placing an order? (Hints: Expoential distribution) A. 0.48658 B. 0.22313 C. 0.51342 D. 0.77687

Respuesta :

Answer:

Step-by-step explanation:

[tex]m = \frac{1}{3}[/tex]

Probability of more than time more than 4.5 minutes

= [tex]e^{-mx}[/tex]

[tex]m = \frac{1}{3}[/tex]

x = 4.5

probability required.

=  [tex]e^{-\frac{1}{3}\times 4.5}[/tex]

=[tex]e^{-1.5}[/tex]

= .22313

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