Normal curve area and z-score explorer

A z-score is a change of units, not a change of distribution. Drag a boundary below and the raw value, the z-score, the shaded probability, and the percentile all move as one thing.

P(85 < X < 115) = 0.6827shaded 68.27%lower 85 · z -1.00 · percentile 15.9upper 115 · z 1.00 · percentile 84.1
meanxz55-370-285-1100011511302145385z = -1.00115z = 1.00
drag a boundary, or click anywhere on the curve to send the nearest one there · arrow keys nudge, shift for finethe curve is drawn from mean minus 4 SD to mean plus 4 SD, so its shape never changes

One SD either side of the mean, from 85 to 115, holds 0.6827 of the curve. That is the 68 in the 68-95-99.7 rule, and it stays 0.6827 for every mean and every SD you can set.

Try to: shade about 95% between two symmetric bounds find the 90th percentile with the right tail push a boundary past z = 2 lock the z-scores, then move the mean

What to try

  1. Read both axes at once. The row under the curve gives the raw value, and the row under that gives its z-score. Drag a boundary and watch a value like 115 and a z-score like 1.00 stay welded together. That pairing is the entire skill this topic asks for.
  2. Lock the z-scores, then move the mean. Turn the lock on and drag the mean slider from 100 to 160. Every number on the value axis changes and the shaded probability does not move at all. One z-score always cuts off one area, on every normal distribution there is.
  3. Measure the empirical rule instead of memorizing it. Choose "between", set the bounds one standard deviation either side of the mean, and read 0.6827. Push them out to two SDs for 0.9545 and three for 0.9973. Those are the real numbers behind 68, 95, and 99.7.
  4. Find where 1.96 comes from. Still in "between", nudge the symmetric bounds until the area reads exactly 0.9500. The z-scores land on plus and minus 1.96, which is the number every 95% confidence interval multiplies by.
  5. Work backwards from a percentile. Switch to the right tail and drag until the shaded area is 0.10. Your boundary is now the 90th percentile, which is the invNorm question in reverse.

Notice that the drawn curve never changes shape. The window is always the mean plus or minus four standard deviations, so every normal distribution looks identical once you measure the horizontal axis in standard deviations. That is what standardizing buys you, and it is why a single z-table serves every normal distribution you will ever meet.

The arithmetic behind the boundary line is in how to find a z-score, and the three landmark areas are in the empirical rule. When you want a typed answer rather than a dragged one, use the normal distribution calculator or look the area up by hand in the z-table. On a graphing calculator the same job belongs to normalcdf and invNorm.

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