Normal Distribution Calculator

Step-by-step solution

1. 1

   Standardize to a z-score

   Subtract the mean μ = 100 from x = 120 and divide by the standard deviation σ = 15. All inputs are rational, so the z-score is an exact fraction: z = 4/3 ≈ 1.333333.

   $$z = \frac{x - \mu}{\sigma} = \frac{120 - 100}{15} = \frac{4}{3} \approx 1.333333$$
2. 2

   Evaluate Φ(z)

   Φ(z) = ½·(1 + erf(z/√2)) has no closed form in elementary functions, so a numerical approximation of the error function is used: Φ(1.333333) ≈ 0.908789.

   $$\Phi\left(\frac{4}{3}\right) \approx 0.908789$$
3. 3

   Interpret the result

   About 90.8789% of the values of a N(100, 15²) distribution lie below x = 120.

   $$P(X < 120) \approx 0.908789 \; (90.8789\%)$$
4. 4

   Result

   P(X ≤ 120) ≈ 0.908789, i.e. about 90.8789%, for X ~ N(100, 15²).

   $$P(X < 120) \approx 0.908789$$