Z = (score-mean)/SD
√Variance = SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion and its Z score.
Put the resulting values in the above equation and solve for the score.
a raw score on a particular intelligence test is known to be approximately normally distributed in a population of 4th graders, with a mean of 95 and variance 81. which score has the property that there are 15% of scores in the population that are above it?
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