1…Using Chebyshev’s inequality, what is the minimum proportion of observations from a population of 500 that must lie within two standard deviations of the mean, regardless of the shape of the distribution? A. 71%. B. 75%. C. 89%. D. 99%. Solution? 1-1/k^2 ?? I still do not get any of the mutiple choice answers… 2…If a distribution exhibits positive skewness, then the mean most likely is located to the: A. left of both the median and mode. B. right of both the median and mode. C. left of the median and right of the mode. D. right of the median and left of the mode. Shouldnt the answer be “B” ???

B&B 1 - 1/2^2 = .75 Just draw a +vely skewed distribution.

2 should be B. I think its a bs answer and I lost it when I saw it was wrong. Is there errata present about this? 1. If you look in the book, 75% is the listed inequality for ± 2 standard deviations from the mean.

1…: 1 - 1/k^2 = 1 - 1 / 2^2 = .75 or 75% => B is your answer. 2…: B

B B Chebyshev’s formula for the first, the second I just draw it out…the mean is always on one side of both the median and mode, unless its a normal distribution.