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Posted by on Sunday, March 17, 2013 at 10:22pm.

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 - , Monday, March 18, 2013 at 8:11pm

    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 - , Sunday, December 6, 2015 at 5:03pm

    43.68

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