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1 December, 22:26

Assume a normal distribution and that the average phone call in a certain town lasted 9 min, with a standard deviation of 1 min. What percentage of the calls lasted less than 8 min?

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Answers (2)
  1. 1 December, 22:42
    0
    The percentage of the calls lasted less than 8 min is 16%.

    Step-by-step explanation:

    We are dealing with a normal distribution with an average phone call of 9 min and a standard deviation of 1 min. Below we can observe the empirical rule applied with a mean of 9 and a standard deviation of 1. The number 8 represents one standard deviation below the mean, so, the percentage of observations below 8 is 16%. Therefore the percentage of the calls lasted less than 8 min is 16%.
  2. 2 December, 00:09
    0
    The percentage of the calls lasted less than 8 min is 16%

    Step-by-step explanation:

    * Lets explain how to solve the problem

    - To find the percentage of the calls lasted less than 8 min, find the

    z-score for the calls lasted

    ∵ The rule of z-score is z = (x - μ) / σ, where

    # x is the score

    # μ is the mean

    # σ is the standard deviation

    * Lets solve the problem

    - The average phone call in a certain town lasted is 9 min

    ∴ The mean (μ) = 9

    - The standard deviation is 1 min

    ∴ σ = 1

    - The calls lasted less than 8 min

    ∴ x = 8

    ∵ z = (x - μ) / σ

    ∴ z = (8 - 9) / 1 = - 1/1 = - 1

    ∴ P (z < 8) = - 1

    - Use z-table to find the percentage of x < 8

    ∴ P (x < 8) = 0.15866 * 100% = 15.87% ≅ 16%

    * The percentage of the calls lasted less than 8 min is 16%
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