Translate between raw units and standard-deviation units

1. 1
   Center

   Find the signed distance from the mean.

   $$x-\mu=86-74=12$$
2. 2
   Scale

   Express the distance in six-unit standard deviations.

   $$z=12/6=2$$

Independent check: The inverse relation 74 + 2(6) returns 86.

1. 1
   Find the signed deviation

   The observation is nine units below the mean.

   $$41-50=-9$$
2. 2
   Standardize

   Divide by the positive standard deviation.

   $$z=-9/3=-3$$

Independent check: Recovering x gives 50 + (-3)(3) = 41.

1. 1
   Subtract

   The raw value and mean match.

   $$x-\mu=18-18=0$$
2. 2
   Divide

   Zero divided by a positive number remains zero.

   $$z=0/2.5=0$$

Independent check: The inverse formula 18 + 0(2.5) returns 18.

1. 1
   Scale the standardized distance

   Convert SD units back to raw units.

   $$z\sigma=(-1.25)(8)=-10$$
2. 2
   Add the mean

   Move ten units below 64.

   $$x=64-10=54$$

Independent check: Forward standardization gives (54 - 64)/8 = -1.25.

1. 1
   Standardize A

   A is 12 units above a six-unit SD.

   $$z_A=(82-70)/6=2$$
2. 2
   Standardize B

   B is 40 units above a 25-unit SD.

   $$z_B=(640-600)/25=1.6$$
3. 3
   Compare

   The greater z-score is farther above its group mean.

   $$2>1.6$$

Independent check: Reverse-standardizing both scores reproduces 82 and 640 with their respective scales.

1. 1
   Rearrange

   Isolate the mean.

   $$\mu=x-z\sigma$$
2. 2
   Substitute

   The standardized displacement is 12.

   $$\mu=72-(1.5)(8)=60$$

Independent check: Using μ = 60 in (72 - μ)/8 gives z = 1.5.

1. 1
   Find the raw deviation

   The value is ten units above the mean.

   $$x-\mu=10$$
2. 2
   Divide by z

   Two standard deviations span those ten units.

   $$\sigma=10/2=5$$

Independent check: Substitution gives (35 - 25)/5 = 2.

1. 1
   Transform center and spread

   The mean shifts and scales; SD only scales.

   $$\mu_Y=1.8\mu_X+32,\quad\sigma_Y=1.8\sigma_X$$
2. 2
   Simplify z

   The shift cancels and the scale factor divides out.

   $$\frac{Y-\mu_Y}{\sigma_Y}=\frac{1.8(X-\mu_X)}{1.8\sigma_X}=z_X$$

Independent check: A value one SD above the mean remains one transformed SD above the transformed mean.

1. 1
   Read the sign

   Positive means above the reference mean.

   $$z=1>0$$
2. 2
   Read the magnitude

   The distance is one reference standard deviation.

   $$x=\mu+\sigma$$

Independent check: Distributions with different shapes can assign different percentile ranks to the same z-score.

1. 1
   Convert to raw units

   Scale the standardized distance.

   $$z\sigma=(1.96)(10)=19.6$$
2. 2
   Locate the cutoff

   Add the displacement to the mean.

   $$x=100+19.6=119.6$$

Independent check: Forward calculation gives (119.6 - 100)/10 = 1.96.