Z-score of an IQ of 130

Question

IQ scores are modeled as X \sim N(	ext{mean } 100,\ 	ext{variance } 225), so the standard deviation is \sqrt{225} = 15. What is the z-score (standardized value) of an IQ of 130?

Accepted Answer

Standardizing is a special linear transform that recenters and rescales any normal into a standard normal N(0,1): subtract the mean, then divide by the standard deviation, Z = \frac{X - \mu}{\sigma}. This is exactly aX + b with a = 1/\sigma and b = -\mu/\sigma, chosen so the new mean is 0 and the new variance is 1. Plugging in: z = \frac{130 - 100}{15} = \frac{30}{15} = 2. So an IQ of 130 sits 2 standard deviations above the mean. Takeaway. The z-score (X-\mu)/\sigma measures distance from the mean in standard-deviation units; here 130 is exactly 2 above.

Z-score of an IQ of 130

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