Finding Probabilities and Values from a Normal Model

For normal X with mean μ and positive standard deviation σ, use z = (x - μ) / σ. A left-cumulative table lists Φ(z) = P(Z < z). Read its stated z row and column; do not use a mean-to-z table as if it listed left areas. The questions here include the table entries you need. With μ = 46, σ = 3 and x = 49, z = 1. Since Φ(1) = 0.8413, P(X < 49) = 0.8413. Symmetry gives Φ(0) = 0.5.

For a normal model, about 68%, 95% and 99.7% lie within one, two and three σ of μ. The central 95% leaves about 5% outside. Symmetry splits that into about 2.5% per tail. Thus an upper 2.5% cutoff is about μ + 2σ, and a lower 2.5% cutoff is about μ - 2σ. With μ = 40 and σ = 6, these approximate cutoffs are 52 and 28.

For an upper tail, subtract the left area from 1. For a < X < b, standardise both ends and use Φ(z_b) - Φ(z_a). The area below b includes the interval and everything below a. With μ = 20 and σ = 2, boundaries 19 and 23 give z = -0.5 and 1.5. Using Φ(-0.5) = 0.3085 and Φ(1.5) = 0.9332 gives 0.6247.

To find a cutoff from a probability, first express the area to its left. Search that area among the left-cumulative table entries, then read the corresponding z. Reverse with x = μ + zσ. For the top 10%, the left area is 0.90 and z is about 1.28. With μ = 80 and σ = 5, the cutoff is about 80 + 1.28 × 5 = 86.4. Lower-tail cutoffs use negative z when their left area is below 0.5.

A given left probability supplies z through the left-cumulative table. Then solve x = μ + zσ for the unknown parameter: μ = x - zσ or σ = (x - μ) / z when z is not 0. For P(X < 35) = 0.8413 and σ = 4, z = 1 gives μ = 31. Two known quantiles also work: b - a = (z_b - z_a)σ, so divide their raw gap by their z gap to find σ.