# 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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