TR

Normal distribution and Z score

Enter the mean and standard deviation to find the probability of falling below, above or between values. The steps also show how to read the result from a table.

Example:

Plant height in a maize variety is normally distributed with mean 210 cm and standard deviation 15 cm. What is the probability that a random plant is taller than 240 cm?

-4-3-2-101234165180195210225240255z = 2Top row z, bottom row original scale

The shaded area is the probability. The total area under the curve is 1.

Result

P(X>240)P(X > 240) = 0.0228

  1. First convert the value(s) to Z scores: how many standard deviations from the mean?

  2. z=x−μσ=240−21015=2z = \dfrac{x - \mu}{\sigma} = \dfrac{240 - 210}{15} = 2

  3. Reading the table (z = 2): 0.9772 in the left-tail table, 0.4772 in the (0, z) table.

  4. The area on the right is 1 (the total area) minus the area on the left.

    P(Z>2)=1−0.9772=0.0228P(Z > 2) = 1 - 0.9772 = 0.0228

  5. As a percentage: 2.28%.

T score of the same value: 50 + 10·z = 70. T scores avoid negative numbers; they have mean 50 and standard deviation 10.