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Normal Distribution worksheet

Grades 11–12 · CCSS HSS.ID.A.4

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- What do you want to find?: Probability below or above x
- Mean (μ): 100
- Standard deviation (σ): 13
- Value (x): 110
Find: P(X < 110)
Answer
2.
Problem 2.
- What do you want to find?: Value for a probability (inverse)
- Mean (μ): 100
- Standard deviation (σ): 18
- Cumulative probability, P(X < x): 0.78
Find: x with P(X < x) = 0.78
Answer
3.
Problem 3.
- What do you want to find?: Value for a probability (inverse)
- Mean (μ): 82
- Standard deviation (σ): 16
- Cumulative probability, P(X < x): 0.9
Find: x with P(X < x) = 0.9
Answer
4.
Problem 4.
- What do you want to find?: Value for a probability (inverse)
- Mean (μ): 110
- Standard deviation (σ): 15
- Cumulative probability, P(X < x): 0.9
Find: x with P(X < x) = 0.9
Answer
5.
Problem 5.
- What do you want to find?: Value for a probability (inverse)
- Mean (μ): 90
- Standard deviation (σ): 14
- Cumulative probability, P(X < x): 0.9
Find: x with P(X < x) = 0.9
Answer
6.
Problem 6.
- What do you want to find?: Value for a probability (inverse)
- Mean (μ): 90
- Standard deviation (σ): 13
- Cumulative probability, P(X < x): 0.9
Find: x with P(X < x) = 0.9
Answer
7.
Problem 7.
- What do you want to find?: Probability below or above x
- Mean (μ): 140
- Standard deviation (σ): 12
- Value (x): 150
Find: P(X < 150)
Answer
8.
Problem 8.
- What do you want to find?: Probability below or above x
- Mean (μ): 80
- Standard deviation (σ): 14
- Value (x): 140
Find: P(X < 140)
Answer
9.
Problem 9.
- What do you want to find?: Probability between a and b
- Mean (μ): 100
- Standard deviation (σ): 12
- Lower bound (a): 65
- Upper bound (b): 110
Find: P(65 < X < 110)
Answer
10.
Problem 10.
- What do you want to find?: Probability below or above x
- Mean (μ): 120
- Standard deviation (σ): 13
- Value (x): 125
Find: P(X < 125)
Answer

Answer key: Normal Distribution worksheet

1. P(X < 110): 0.78
1. Standardize: turn the value into a z-score, the number of standard deviations it sits from the mean, so one standard table works for any bell curve
  $z$ $= \frac{x - μ}{σ}$
2. Put in your value, mean and standard deviation
  $z$ $= \frac{110 - 100}{13}$ $= 0.7692307692$
3. Look up Φ(z), the share of the standard bell curve to the left of z. That's the chance of a value below x
  $P(X < 110)$ $= Φ(0.769231)$ $= 0.7791218363$
4. The whole curve adds up to 1, so the area to the right is 1 minus the area to the left
  $P(X > 110)$ $= 1 - 0.7791218363$ $= 0.2208781637$
2. x with P(X < x) = 0.78: 113.9
1. Work backwards: find the z-score that has your probability to its left (the inverse of Φ)
  $z$ $= Φ⁻¹(0.78)$ $= 0.7721932142$
2. Undo the standardizing to get back to your scale: start at the mean μ and add z standard deviations σ
  $x$ $= μ + z \times σ$
3. Put in your mean, z-score and standard deviation
  $x$ $= 100 + 0.7721932142 \times 18$ $= 113.8994778554$
3. x with P(X < x) = 0.9: 102.5
1. Work backwards: find the z-score that has your probability to its left (the inverse of Φ)
  $z$ $= Φ⁻¹(0.9)$ $= 1.2815515655$
2. Undo the standardizing to get back to your scale: start at the mean μ and add z standard deviations σ
  $x$ $= μ + z \times σ$
3. Put in your mean, z-score and standard deviation
  $x$ $= 82 + 1.2815515655 \times 16$ $= 102.5048250487$
4. x with P(X < x) = 0.9: 129.22
1. Work backwards: find the z-score that has your probability to its left (the inverse of Φ)
  $z$ $= Φ⁻¹(0.9)$ $= 1.2815515655$
2. Undo the standardizing to get back to your scale: start at the mean μ and add z standard deviations σ
  $x$ $= μ + z \times σ$
3. Put in your mean, z-score and standard deviation
  $x$ $= 110 + 1.2815515655 \times 15$ $= 129.2232734832$
5. x with P(X < x) = 0.9: 107.94
1. Work backwards: find the z-score that has your probability to its left (the inverse of Φ)
  $z$ $= Φ⁻¹(0.9)$ $= 1.2815515655$
2. Undo the standardizing to get back to your scale: start at the mean μ and add z standard deviations σ
  $x$ $= μ + z \times σ$
3. Put in your mean, z-score and standard deviation
  $x$ $= 90 + 1.2815515655 \times 14$ $= 107.9417219176$
6. x with P(X < x) = 0.9: 106.66
1. Work backwards: find the z-score that has your probability to its left (the inverse of Φ)
  $z$ $= Φ⁻¹(0.9)$ $= 1.2815515655$
2. Undo the standardizing to get back to your scale: start at the mean μ and add z standard deviations σ
  $x$ $= μ + z \times σ$
3. Put in your mean, z-score and standard deviation
  $x$ $= 90 + 1.2815515655 \times 13$ $= 106.6601703521$
7. P(X < 150): 0.8
1. Standardize: turn the value into a z-score, the number of standard deviations it sits from the mean, so one standard table works for any bell curve
  $z$ $= \frac{x - μ}{σ}$
2. Put in your value, mean and standard deviation
  $z$ $= \frac{150 - 140}{12}$ $= 0.8333333333$
3. Look up Φ(z), the share of the standard bell curve to the left of z. That's the chance of a value below x
  $P(X < 150)$ $= Φ(0.833333)$ $= 0.797671619$
4. The whole curve adds up to 1, so the area to the right is 1 minus the area to the left
  $P(X > 150)$ $= 1 - 0.797671619$ $= 0.202328381$
8. P(X < 140): 1
1. Standardize: turn the value into a z-score, the number of standard deviations it sits from the mean, so one standard table works for any bell curve
  $z$ $= \frac{x - μ}{σ}$
2. Put in your value, mean and standard deviation
  $z$ $= \frac{140 - 80}{14}$ $= 4.2857142857$
3. Look up Φ(z), the share of the standard bell curve to the left of z. That's the chance of a value below x
  $P(X < 140)$ $= Φ(4.285714)$ $= 0.9999908924$
4. The whole curve adds up to 1, so the area to the right is 1 minus the area to the left
  $P(X > 140)$ $= 1 - 0.9999908924$ $= 0.0000091076$
9. P(65 < X < 110): 0.8
1. Standardize: turn the value into a z-score, the number of standard deviations it sits from the mean, so one standard table works for any bell curve
  $z$ $= \frac{x - μ}{σ}$
2. Turn the lower bound a into a z-score
  $z_a$ $= \frac{65 - 100}{12}$ $= −2.9166666667$
3. Turn the upper bound b into a z-score
  $z_b$ $= \frac{110 - 100}{12}$ $= 0.8333333333$
4. Take the area up to b and subtract the area up to a. What's left is the share between them
  $P(65 < X < 110)$ $= Φ(0.833333) - Φ(-2.916667)$ $= 0.797671619 - 0.0017689682$ $= 0.7959026508$
10. P(X < 125): 0.65
1. Standardize: turn the value into a z-score, the number of standard deviations it sits from the mean, so one standard table works for any bell curve
  $z$ $= \frac{x - μ}{σ}$
2. Put in your value, mean and standard deviation
  $z$ $= \frac{125 - 120}{13}$ $= 0.3846153846$
3. Look up Φ(z), the share of the standard bell curve to the left of z. That's the chance of a value below x
  $P(X < 125)$ $= Φ(0.384615)$ $= 0.6497388029$
4. The whole curve adds up to 1, so the area to the right is 1 minus the area to the left
  $P(X > 125)$ $= 1 - 0.6497388029$ $= 0.3502611971$