Normal-Distribution Reliability Conversion

Content

Estimate the percentage of a normally distributed life population that survives beyond a selected service life.

Engineering screening tool—verify inputs and results independently. Use at your own risk.

Life-distribution inputs

Use one consistent time unit and a positive normal-distribution mean and standard deviation.

tμR(t): survival area

Modeled reliability

Reliability R(t)89.98%modeled population surviving beyond t
Standardized margin z1.280
Failure probability by t10.02%

Positive z places service life below the mean. Reliability is evaluated with the finite standard-normal density sum documented below.

How to use

  1. Enter positive mean life μ and standard deviation σ in hours, then enter non-negative service life t in hours. All three fields must share one time unit.
  2. Select Calculate or edit any field to update the result. Select Reset to restore the worked example.
  3. Read R(t) as the modeled percentage surviving beyond t. The standardized margin z shows how far t lies below or above μ in units of σ.
  4. Correct any warning before use. The supported integration range is −5 ≤ z ≤ 5.
  5. Do not apply this model to non-normal, censored, multimodal, repairable, dependent, or changing populations, or to safety and certification decisions without validated reliability analysis.

Equations used

Life and scatter are positive magnitudes; service time is non-negative. The implemented equations are:

z = (μ − t)/σ
φ(x) = exp(−x²/2)/√(2π)
Δz = 0.001; xᵢ = −5 + iΔz
k = 1000(z + 5)
Fraw = Σᵢ₌₁ᵏ φ(xᵢ)Δz
R(t) = 100 × round(Fraw × 10000)/10000 [%]

μ is mean life, σ is the life standard deviation, and t is service life, all in hours. z and Fraw are dimensionless. The sum uses right endpoints and the loop condition i < k + 1; this retains fractional-endpoint behavior. The displayed percent is equivalently Math.round(Fraw × 10000)/100, or two decimal places.

Worked example

  1. Enter μ = 6000 h, σ = 1250 h, and t = 4400 h.
  2. The standardized margin is z = (6000 − 4400)/1250 = 1.280.
  3. The finite density sum gives R(t) = 89.98%. About 89.98% of the modeled population survives beyond 4400 h; about 10.02% does not.

Theory and method

Standard normal density

A normal life model represents population life with mean μ and standard deviation σ. Standardizing the service threshold produces z = (μ − t)/σ, so the survival probability beyond t equals the standard-normal cumulative probability through z. This calculator evaluates that probability by summing the standard normal density from −5 to z in increments of 0.001, preserving the implemented finite numerical method and percent rounding.

Assumptions and limits

  • Population life is adequately normal with stable μ and σ, and all values use one time unit.
  • Only −5 ≤ z ≤ 5 is accepted. The finite lower tail and 0.001 step are approximations, not an exact modern normal CDF.
  • A normal distribution can imply negative life when σ is large relative to μ; use a non-negative life distribution when that implication is material.
  • Censoring, mixtures, degradation, repairs, dependencies, confidence intervals, and parameter uncertainty are omitted.

References

Related analysis: Series and Parallel System Reliability Calculator. Continue in TriboSolver when physics-based loads, contact, lubrication, or wear should refine the life assumptions before reliability screening.