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Statistics

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32% of adult internet users have purchased products or services online. For a random sampling of 200 adult users, find the mean, variance and standard deviation for the number who have purchased goods or services online?

  • Statistics -

    Mean = np = (200)(.32) = ?

    Variance = npq = (200)(.32)(.68) = ?
    (Note: q = 1 - p)

    Standard deviation = square root of the variance

    I'll let you finish the calculations.

  • Statistics -

    43.68

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