SL 4.12—Standardizing normal variables

Standardisation measures distance from the mean in standard deviations.

For X with mean μ and standard deviation σ, z=(X−μ)/σ lets normal probabilities be read on a common scale.

Example

A score 85 from N(70,10²) has z=1.5, so it is 1.5 standard deviations above the mean.

Keep the sign: negative z-values lie below the mean, and convert back with X=μ+σz.

A z-score is relative position, not a percentage or a guarantee of rarity.

Unknown-parameter example: suppose a normal model assigns $z=-1$ to $x=70$ and $z=2$ to $x=100$. Then $(70-\mu)/\sigma=-1$ and $(100-\mu)/\sigma=2$. Subtracting gives $30/\sigma=3$, so $\sigma=10$ and then $\mu=80$. Use technology to obtain the needed inverse-normal cut-offs, then solve the resulting z-equations with $\sigma>0$.