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13 February, 18:23

A travel agency did a survey and found that the average local family spends $1900 on a summer vacation. The distribution is normally distributed with standard deviation $390. A. What percent of the families took vacations that cost under $1500. Round to the nearest percent.

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  1. 13 February, 20:49
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    The percentage is 15%

    Step-by-step explanation:

    * Lets revise how to find the z-score

    - The rule the 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 local family spends $1900 on a summer vacation

    - The distribution is normally distributed with standard deviation $390

    - We need to find what percent of the families took vacations that

    cost under $1500

    ∵ The mean is $ 1900

    ∴ μ = 1900

    ∵ The standard deviation is $390

    ∴ σ = 390

    ∵ The vacation cost is under $1500

    ∴ x = 1500

    ∵ z-score = (x - μ) / σ

    - Substitute the values above in the rule

    ∴ z-score = (1500 - 1900) / 390 = - 1.0256

    - Lets use the normal distribution table of z

    ∵ P (z < 1500) = 0.1515

    ∵ P (x < 1500) = P (z < 1500)

    ∴ P (x < 1500) = 0.1515

    ∴ The percentage of the families took vacations that cost under

    $1500 is ⇒ 0.15 * 100% = 15%

    ∴ The percentage is 15%
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