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13 December, 17:58

The customer support department for a major computer supplier believes that the time between consecutive phone inquiries is exponentially distributed, with an average value of 60 seconds.

Part a What is the probability that the time between two consecutive phone inquiries is less than 50 seconds?

Part b Consider a random selection of 100 consecutive phone inquiries. What is the probability that the average time between consecutive inquiries in this random sample is less than 50 seconds?

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  1. 13 December, 18:10
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    a) P (X < 50) = 0.56540

    b) P (x < 50) = 0.1193

    Step-by-step explanation:

    Given:-

    - The average time between consecutive phone inquiries = 60 seconds

    Find:-

    a What is the probability that the time between two consecutive phone inquiries is less than 50 seconds?

    b Consider a random selection of 100 consecutive phone inquiries. What is the probability that the average time between consecutive inquiries in this random sample is less than 50 seconds?

    Solution:-

    - A random variable (X) follows an exponential distribution with rate scale parameter λ = 1 / average = 1/60.

    X ~ Exp (1 / 60)

    a)

    - The CDF of exponential distribution is given by:

    P (X < x) = 1 - e^ (-λx)

    P (X < 50) = 1 - e^ (-50/60)

    = 0.56540

    b)

    - A sample of n = 100 consecutive phone inquiries was taken to determine probability that the average time between consecutive inquiries in this random sample is less than 50 seconds.

    - We will approximate the sample taken to be normally distributed. The mean (u) and standard deviation (s) of the normally distributed sample would be:

    u = 1 / λ = 60 seconds

    s = 1 / √n*λ = 60 / √50 = 8.48528

    Then,

    x ~ N (60, 8.48528^2)

    - We will standard our test value:

    P (x < 50) = P (z < (50 - 60) / 8.48528)

    P (z < - 1.17851) = 0.1193

    - The result is P (x < 50) = 0.1193
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