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The average life of Canadian women is 73.75 years and the standard deviation of the women's life expectancy in Canada is 6.5 years. Using the Chebyshev's theorem, determine the minimum percentage of women in Canada whose life expectancy is between 64 and 83.5 years. Round your answer to 2 decimal places and express as a percentage (%).

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    Some hints:

    This theorem says
    1. Within two standard deviations of the mean, you will find at least 75% of the data.
    2. Within three standard deviations of the mean, you will find at least 89% of the data.

    Here's how the formula shows this:
    Formula is 1 - (1/k^2) ---> ^2 means squared.

    If k = 2 (representing two standard deviations), we have this:
    1 - (1/2^2) = 1 - (1/4) = 3/4 or .75 or 75%

    If k = 3 (representing three standard deviations), we have this:
    1 - (1/3^2) = 1 - (1/9) = 8/9 or approximately .89 or 89%

    Looking at your problem, the difference between the mean and the data given is 1.5 standard deviations.
    Therefore, we have this:
    1 - (1/1.5^2) = 1 - (1/2.25) = 1 - 0.44 = 0.56 = 56%

    I hope this helps.

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