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Thus based on the data in the sample an sich (as calculated by a test for normality), it is unlikely that the sample is part of a normal distribution, but we already know it is part of a normal distribution, because it's part of population that is normally distributed.
Zenia Wilson 3/8/2013 6.89 A variable is normally distributed with mean 6 and standard deviation 2. a. Determine and interpret the quartiles of the variable. Find the x-value having the z-score -0.67 x = + z = 6 + 2(-0.67) = -4.65 the first quartile is -4.65. x = + z = 6 + 2(0) = 6 the second quartile is 6. x = + z = 6 + 2(-0.67) = -4.65 the third quartile is 7.35 b. Obtain and interpret the 85th percentile.