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If a distribution of intelligence quotient (IQ) scores has a mean of $$100$$ and a standard deviation of $$15$$, an IQ score of $$110$$ would have a $$z$$ score of $$\frac{110 - 100}{15} = +0.67$$. This means the score is approximately two-thirds of a standard deviation above the mean. Conversely, a raw score of $$85$$ would have a $$z$$ score of $$\frac{85 - 100}{15} = -1.00$$, placing it exactly one standard deviation below the mean.

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