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.
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?
2 answers
43.68