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28 August, 22:04

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 is the probability that a randomly selected caller is placed on hold fewer than 7 minutes?

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  1. 29 August, 01:29
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    0.917915

    Step-by-step explanation:

    Exponential distribution is expressed as it follows

    F (y,λ) = 1 - e ^ ( - λ (y))

    Where λ is equal to 1 / population mean

    So λ = 1/2.8 = 0.357

    Let y be the length of time a costumer wait on hold

    Now, let's find the value for y = 7 minutes, which is the proportion of costumer that wait on hold

    P (y<7) = 1 - e^ (-.357 * (7))

    P (y<7) = 0.917915
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