Weibull Distribution Calculator

Find Weibull distribution probabilities, reliability function, mean, median, and full distributional statistics for any shape and scale parameters.

🔧 Weibull Distribution Calculator
Shape parameter k1.5
0.110
Scale parameter λ1
0.110
Query value x1
020
Shape parameter k2
0.110
Scale parameter λ1
0.110
CDF — P(X ≤ x)
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Reliability — S(x)
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PDF f(x)
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Mean (μ)
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Variance
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Std Deviation
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Median
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Mode
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Mean (μ)
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Median
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Mode
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Variance
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Std Deviation
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Skewness
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Excess Kurtosis
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🔧 What is the Weibull Distribution?

The Weibull distribution is a versatile continuous probability distribution widely used in reliability engineering, survival analysis, and extreme value theory. Named after Swedish mathematician Waloddi Weibull who described it in 1951, it is exceptionally useful because its shape can model increasing, constant, or decreasing failure rates depending on the shape parameter k (also called beta). No other two-parameter distribution covers this range of behaviors so compactly.

The distribution has two parameters: the shape parameter k (dimensionless) and the scale parameter lambda (same units as the variable). Three regimes arise from k: when k is less than 1, the hazard rate (instantaneous failure rate) decreases with time, modeling infant mortality or early-life failures due to manufacturing defects. When k equals 1, the hazard is constant, reducing the Weibull to the exponential distribution and modeling random failures independent of age. When k is greater than 1, the hazard increases, modeling wear-out or aging failures. Wind turbine blades, semiconductor lifetimes, bearing wear, and material fatigue all follow Weibull distributions with characteristic k values fitted from field data.

A common misconception is that the Weibull distribution is only for failure-time data. In practice, it is used across many fields: wind speed distributions (k near 2, the Rayleigh special case), ocean wave heights, tensile strength of materials, particle size distributions in milling operations, and even extreme rainfall events. The key requirement is that the physical process produces a positively skewed distribution bounded below by zero, which the Weibull handles for a very wide range of shapes through its single k parameter.

The scale parameter lambda deserves special attention. Regardless of k, the CDF always equals 1 minus 1/e, approximately 63.21%, when x equals lambda. This makes lambda the "characteristic life" or "eta" in reliability terminology: 63.21% of the population has failed by time lambda. Engineers often report lambda and k as the two key descriptors of a component's life distribution, fitting them from time-to-failure data using maximum likelihood or least squares on a Weibull probability plot (Weibull plot).

📐 Formula

f(x; k, λ)  =  (k / λ) × (x / λ)k−1 × e−(x/λ)k
f(x) = probability density function at value x
k = shape parameter (k > 0); controls hazard rate trend
λ = scale parameter (lambda > 0); characteristic life (x at 63.21% CDF)
CDF: P(X ≤ x) = 1 − e−(x/λ)k
Reliability: S(x) = e−(x/λ)k
Mean: μ = λ × Γ(1 + 1/k)
Median: λ × (ln 2)1/k
Mode: λ × ((k−1)/k)1/k  (for k > 1; else mode = 0)
Variance: λ² × [Γ(1 + 2/k) − Γ(1 + 1/k)2]
Example: k = 2, λ = 2, x = 2: CDF = 1 − e−1 ≈ 63.21%

📖 How to Use This Calculator

Steps

1
Choose a calculation mode - Select "Calculate Probability" to evaluate the CDF, reliability, and PDF at a specific x value, or select "Distribution Stats" for the full parameter set including skewness and kurtosis.
2
Enter the shape parameter k - Type k or drag the slider. Use k less than 1 for infant mortality, k = 1 for exponential, k = 1.5-2 for moderate wear-out, k near 3.6 for near-normal. Common reliability engineering values are k = 1 to k = 4.
3
Enter the scale parameter lambda - Set lambda to the characteristic life of your population. The CDF equals 63.21% at x = lambda for any k. Use the same units as your data (hours, cycles, km).
4
Enter the query value x (Probability mode) - Set x to the time or value at which you want the CDF and reliability. Mean, median, mode, variance, and SD are also displayed for the given k and lambda.

💡 Example Calculations

Example 1 - Bearing Wear-out (k = 2, lambda = 1000 h, x = 800 h)

A bearing has Weibull failure distribution with shape k=2 and characteristic life lambda=1000 h. What fraction fail by 800 h?

1
CDF = 1 - exp(-(800/1000)^2) = 1 - exp(-0.64) = 1 - 0.5273 = 0.4727 = 47.27%.
2
Reliability S(800) = exp(-0.64) = 52.73%, meaning 52.73% of bearings survive past 800 hours.
3
Mean life = 1000 * Gamma(1.5) = 1000 * sqrt(pi)/2 = 886.2 h. Median = 1000*(ln2)^0.5 = 832.6 h.
P(failure by 800 h) = 47.27%
Try this example (scaled) →

Example 2 - Early-Life Failures (k = 0.5, lambda = 1, x = 2)

Electronic component with decreasing failure rate (k=0.5, lambda=1). What is P(X at most 2)?

1
(x/lambda)^k = (2/1)^0.5 = sqrt(2) = 1.4142.
2
CDF = 1 - exp(-1.4142) = 1 - 0.2431 = 0.7569 = 75.69%.
3
Note: k less than 1 means the hazard rate decreases over time. Components that survive early are less likely to fail later.
P(X ≤ 2) = 75.69%
Try this example →

Example 3 - Wind Speed Distribution Stats (k = 2, lambda = 7 m/s)

Wind speed at a site follows Weibull(k=2, lambda=7 m/s). Find mean, median, mode, and variance.

1
Mean = 7 * Gamma(1.5) = 7 * sqrt(pi)/2 = 7 * 0.8862 = 6.204 m/s.
2
Median = 7 * (ln2)^0.5 = 7 * 0.8326 = 5.828 m/s. Mode = 7 * (1/2)^0.5 = 7/sqrt(2) = 4.950 m/s.
3
Variance = 49 * [Gamma(2) - Gamma(1.5)^2] = 49 * [1 - pi/4] = 49 * 0.2146 = 10.516 m^2/s^2. SD = 3.243 m/s.
Mean wind speed = 6.204 m/s
Try this example →

❓ Frequently Asked Questions

What is the Weibull distribution used for in reliability engineering?+
The Weibull distribution is the standard tool for modeling component lifetimes and failure-time data. By fitting k and lambda from time-to-failure records, engineers can compute warranty costs, plan preventive maintenance schedules, and certify product reliability. Industries using Weibull analysis include aerospace (turbine blade fatigue), automotive (bearing life), electronics (semiconductor MTTF), and civil engineering (structural load capacity).
How do I estimate the Weibull parameters from data?+
The two main methods are maximum likelihood estimation (MLE) and Weibull probability plot (least squares). For MLE: maximize the log-likelihood ln L = n*ln(k) - n*k*ln(lambda) + (k-1)*sum(ln xi) - sum((xi/lambda)^k). For the Weibull plot method: plot ln(-ln(1 - F_i)) vs ln(x_i) on a log-log scale; the points fall on a line with slope = k and intercept = -k*ln(lambda). Software like R (fitdistr), Python (scipy.stats.weibull_min.fit), and Excel add-ins compute both methods automatically.
What value of k makes the Weibull distribution symmetric?+
The Weibull distribution is approximately symmetric (near-normal shape) when k is near 3.6. At k = 3.6, the skewness is approximately zero and the distribution closely approximates the normal. For k less than 3.6, the distribution is positively skewed (right tail); for k greater than 3.6, it is negatively skewed (left tail). This is why k = 3.6 is sometimes used to model near-normal failure data without assuming a strict normal distribution.
What is the hazard rate of the Weibull distribution?+
The hazard rate (failure rate) is h(x) = f(x)/S(x) = (k/lambda)*(x/lambda)^(k-1). For k less than 1, h(x) decreases with x (infant mortality). For k = 1, h(x) = 1/lambda is constant (exponential). For k greater than 1, h(x) increases with x (wear-out). The hazard rate is the probability of failing in the next instant given survival to time x, expressed per unit time.
How is the Weibull distribution related to the Rayleigh distribution?+
The Rayleigh distribution is a special case of the Weibull distribution with shape k = 2 and scale lambda = sigma*sqrt(2), where sigma is the Rayleigh scale parameter. Setting k=2 in this calculator produces the same CDF as the Rayleigh calculator with sigma = lambda/sqrt(2). This connection appears in wind energy (Rayleigh is the standard model for wind speed, equivalent to Weibull with k=2) and in wireless communications (Rayleigh fading envelope).
What does CDF = 63.21% at x = lambda mean in practice?+
For any Weibull distribution, F(lambda) = 1 - exp(-1) = 63.21%, regardless of the shape k. This makes lambda the "characteristic life" or "B63 life": the age by which 63.21% of the population has failed. Reliability engineers sometimes prefer to specify the B10 life (age at 10% failure) or B50 life (median), which are lambda*(ln(1/0.90))^(1/k) and lambda*(ln2)^(1/k) respectively. The calculator shows the median directly in its output.
How do I calculate the mean time to failure (MTTF) from Weibull parameters?+
MTTF = lambda * Gamma(1 + 1/k), where Gamma is the gamma function. For k=1: MTTF = lambda (exponential mean). For k=2: MTTF = lambda*sqrt(pi)/2 = lambda*0.8862. For k=3: MTTF = lambda*Gamma(4/3) = lambda*0.8930. The MTTF decreases relative to lambda as k increases from 1 toward infinity, while the distribution narrows. This calculator displays the mean automatically from the gamma function approximation.
What is the difference between the Weibull PDF and the hazard function?+
The PDF f(x) describes the unconditional probability density of failure at time x. The hazard function h(x) = f(x)/S(x) is the conditional failure rate given survival to time x. For k greater than 1, h(x) increases over time while f(x) first rises then falls. The PDF peaks at the mode and then decreases; the hazard continues rising (for k greater than 1) past the mode, past the mean, and toward infinity. These are different functions that answer different reliability questions.
Can the Weibull distribution handle negative or zero values?+
The standard two-parameter Weibull distribution (this calculator) is defined only for x at least 0. The PDF equals zero for negative x. A three-parameter Weibull adds a location parameter gamma (threshold below which failure cannot occur), shifting the distribution: x is replaced by x minus gamma. This calculator uses the two-parameter form, which is sufficient for the vast majority of reliability and life-data applications.
How does the Weibull distribution compare to the lognormal for reliability data?+
Both the Weibull and lognormal are used for right-skewed lifetime data. The Weibull has the advantage of a simple closed-form CDF and hazard rate, making it computationally easy and well-suited for censored data analysis. The lognormal often fits fatigue and corrosion data better because the logarithm of time-to-failure is normally distributed. For data where the hazard rate first increases then decreases (bathtub curve after the wear-out phase), neither fits well individually; a mixture model or the Weibull mixture is used instead.
What is the variance and standard deviation of the Weibull distribution?+
The variance is lambda^2 times [Gamma(1+2/k) minus (Gamma(1+1/k))^2]. The standard deviation is the square root of the variance. For k=1 (exponential): variance = lambda^2, SD = lambda. For k=2: variance = lambda^2*(1 - pi/4) = 0.2146*lambda^2, SD = 0.4633*lambda. The coefficient of variation (SD/mean) depends only on k and not on lambda, making it a dimensionless shape descriptor.
How do I use the Weibull distribution for wind energy calculations?+
For wind energy, the Weibull distribution with k near 2 (Rayleigh approximation) and lambda = mean_wind / Gamma(1+1/k) is the standard model for wind speed frequency distributions. Given k and lambda, the energy output of a turbine is integrated as the product of the wind power curve and the Weibull PDF over all wind speeds from cut-in to cut-out. Many wind resource assessment tools (WAsP, WindPro) fit k and lambda from measured data and report annual energy production (AEP) estimates based on the fitted Weibull parameters.
Tags: Weibull Distribution Calculator Weibull Reliability Weibull CDF Weibull PDF Shape and Scale Parameter Calculator Formula Free Online

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What is the Weibull distribution formula?

The PDF is f(x; k, lambda) = (k/lambda)*(x/lambda)^(k-1)*exp(-(x/lambda)^k) for x at least 0. The CDF is F(x) = 1 - exp(-(x/lambda)^k). The reliability function (survival) is S(x) = exp(-(x/lambda)^k). Parameters: k is the shape (dimensionless) and lambda is the scale (same units as x).

What is the mean of the Weibull distribution?

The mean is lambda times Gamma(1 + 1/k), where Gamma is the gamma function. For k=1 (exponential), mean = lambda. For k=2, mean = lambda*sqrt(pi)/2 = lambda*0.8862. For k=3.6, mean is approximately 0.9005*lambda. The mean always exceeds the median for k less than 3.6 (right-skewed) and falls below it for k greater than 3.6.

What does the shape parameter k control in the Weibull distribution?

The shape parameter k (sometimes written beta or c) determines the hazard rate behavior. When k less than 1, the hazard rate decreases over time (infant mortality or early failure period). When k = 1, the hazard is constant (exponential, random failures). When k greater than 1, the hazard increases (wear-out). Values near 3.6 give a symmetric, near-normal shape.

What is the scale parameter lambda in the Weibull distribution?

The scale parameter lambda (also called eta or the characteristic life) is the value at which the CDF equals 1 - 1/e = 63.21%, regardless of the shape k. This is because CDF(lambda) = 1 - exp(-(lambda/lambda)^k) = 1 - exp(-1) = 0.6321. Lambda sets the time scale of the distribution, stretching or shrinking it without changing its shape.

How is the Weibull distribution related to the exponential distribution?

Setting k=1 reduces the Weibull to the exponential distribution with rate 1/lambda. The exponential is the only continuous memoryless distribution. For k not equal to 1, the Weibull is not memoryless, which makes it more realistic for modeling equipment with aging (k greater than 1) or burn-in (k less than 1).

How do I interpret the Weibull reliability function?

The reliability function S(x) = exp(-(x/lambda)^k) gives the probability that a component functions without failure up to time x. For k=2 and lambda=1000 hours, S(800) = exp(-(800/1000)^2) = exp(-0.64) = 0.527, meaning about 52.7% of units survive past 800 hours. Engineers use S(x) to compute MTTF and design maintenance schedules.

What is the median of the Weibull distribution?

The median is lambda times (ln 2)^(1/k), approximately lambda times 0.6931^(1/k). For k=1 (exponential): median = lambda*ln(2) = 0.6931*lambda. For k=2: median = lambda*sqrt(ln2) = lambda*0.8326. The median is less than the characteristic life lambda for all k, because the CDF at x=median is 50% while at x=lambda it is 63.21%.

What is the mode of the Weibull distribution?

For k greater than 1, the mode (most likely value) is lambda*((k-1)/k)^(1/k). For k less than or equal to 1, the distribution is monotonically decreasing and the mode is 0. For k=2 and lambda=1: mode = 1*(1/2)^0.5 = 1/sqrt(2) = 0.7071. For k=5 and lambda=1: mode = (4/5)^(1/5) = 0.8^0.2 = 0.9565.

📌 Quick Tips

💡The shape parameter k determines the failure rate behavior: k less than 1 means decreasing failure rate (infant mortality), k = 1 reduces to the exponential (constant rate), k greater than 1 means increasing failure rate (wear-out failures). Set k near 3.6 to approximate a normal distribution.
💡The scale parameter lambda (also called eta in reliability engineering) is the characteristic life: the time at which 63.21% of units have failed, regardless of the shape k. This is because CDF(lambda) = 1 - exp(-1) = 0.6321 for any k.
💡For reliability work, the survival function S(x) = exp(-(x/lambda)^k) gives the probability that a component survives past time x. This is also called the reliability function R(t) in engineering.
💡The Weibull distribution becomes the exponential distribution when k = 1 (constant hazard rate), the Rayleigh distribution when k = 2 (with scale lambda/sqrt(2)), and approximates the normal when k is near 3.6.
💡Use the Distribution Stats mode to check whether your data looks Weibull-shaped. Positive skewness (k less than 3.6) means the tail extends right; negative skewness (k greater than 3.6) means the tail extends left.