Normal distribution worksheet with answers

Free printable Normal distribution worksheet: 6 problems from easy to exam level, with an answer key. Print it or save it as a PDF.

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1.Easy
−3−2−10123zshaded area to the left of z = 0.0197

Drag z so that the shaded area under the standard normal curve, to the left of z, is 0.0197. Find z to two decimals.

A = 0.0197
z =
2.Easy
32 minσ = ?X + Y

X and Y are independent normal variables for the time for two consecutive process steps: X has mean 12 min and standard deviation 1.5 min, Y has mean 20 min and standard deviation 2 min. Find the standard deviation σ of X + Y.

μX = 12 min, σX = 1.5 minμY = 20 min, σY = 2 min
σ = min
3.Medium
169.95179.75P = ?

The heights of adult men are normally distributed with mean 178 cm and standard deviation 7 cm. What fraction lies between 169.95 and 179.75 cm? Give P as a decimal.

μ = 178σ = 7x1 = 169.95x2 = 179.75
P =
4.Medium
−3−2−10123zshaded area between −z and z = 0.2510

Drag z so that the shaded area between −z and z under the standard normal curve is 0.2510. Find z to two decimals.

A = 0.2510
z =
5.Hard
10 g0X − Y?

The mass X of a product from line 1 and Y from line 2 are independent and normal: X has mean 500 g and standard deviation 8 g, Y has mean 490 g and standard deviation 6 g. Find the probability P that X is greater than Y. Round z to two decimals.

μX = 500 g, σX = 8 gμY = 490 g, σY = 6 g
P =
6.Hard
0.03 mm0D − d?

Hole diameters D are normal with mean 25.03 mm and standard deviation 0.01 mm. Shaft diameters d are independent and normal with mean 25 mm and standard deviation 0.015 mm. A randomly chosen shaft fits when D − d > 0. Find the probability P that it fits. Round z to two decimals.

hole: μ = 25.03 mm, σ = 0.01 mmshaft: μ = 25 mm, σ = 0.015 mm
P =

Answer key

  1. 1. z = −2.06
  2. 2. σ = 2.5 min
  3. 3. P = 0.4736
  4. 4. z = 0.32
  5. 5. P = 0.8413
  6. 6. P = 0.9515

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