What Is a Probability Density Function? (Definition)
The idea of density appears in everyday contexts. When you say a crowd is densely packed at one end of a venue and sparse at the other, you are describing a spatial density — not the exact count at any single point, but the concentration across regions. A PDF works the same way for probability: high f(x) at some value means probability is concentrated there; low f(x) means probability is spread thin. The actual probability of landing in any specific range is the area you measure under the curve, not the height of the curve itself.
This is the single most important fact about PDFs, and it is the source of most conceptual confusion: the value f(x) is not a probability. It is a probability density. To get a probability, you must integrate — that is, you must look at an interval, not a point.
For a continuous random variable X, P(X = x) = 0 for every specific value x. The density f(x) at a point describes concentration of probability near that point, not the probability of hitting that exact point. Probability comes from the integral: P(a ≤ X ≤ b) = ∫[a to b] f(x) dx.
Conditions for a Valid Probability Density Function
A function f(x) qualifies as a probability density function if and only if it satisfies two conditions simultaneously. Both are necessary; satisfying only one is not enough.
The density must be nonnegative everywhere. A negative density would imply negative probability over some interval, which has no sensible interpretation. This condition must hold for all x in the sample space, including values where f(x) = 0.
The entire area under the PDF must equal exactly 1. Because the random variable must take some value, the probability of all possible outcomes together is 1. A function that integrates to anything other than 1 is not a valid PDF.
These two conditions are the complete test. Any function meeting both can serve as the PDF of a continuous random variable. If a function is nonnegative but integrates to 0.8 rather than 1, it is not a valid PDF — though you could divide it by 0.8 to create one. If a function integrates to 1 but dips below zero somewhere, it also fails.
Given f(x) = 2x for 0 ≤ x ≤ 1, and f(x) = 0 otherwise: (1) f(x) = 2x ≥ 0 on [0,1] ✓. (2) ∫[0 to 1] 2x dx = [x²]₀¹ = 1 ✓. Both conditions satisfied — this is a valid PDF.
Probability Density Function Formula
There is no single universal formula for a PDF because different distributions have different functional forms. What is universal is how you calculate probability from any PDF.
f(x) = probability density function
a = lower bound of interval
b = upper bound of interval
∫ = integration (area under the curve)
The formula says: the probability that X lands between a and b is the area under the PDF curve between a and b. You find that area by integrating f(x) from a to b.
Three related formulas follow directly from this definition. The cumulative distribution function (CDF) is the left-tail probability:
The third line is worth pausing on: the PDF is the derivative of the CDF. If you know the CDF, differentiating it recovers the PDF. If you know the PDF, integrating it gives the CDF. They contain the same information, expressed differently.
Probability Density Function Calculator
Select a distribution, enter its parameters, and specify the interval you want. The calculator returns the PDF value at a point, the probability over the interval, the CDF value, and key distribution statistics. All calculations run locally in your browser.
PDF Calculator
Choose a distribution and interval to compute probability. No personal data is collected or transmitted.
PDF Graph and Visualization
The graph below displays the PDF curve for the selected distribution. The shaded region represents the probability over the interval [a, b] you specify — that area is P(a ≤ X ≤ b). The mean is marked with a vertical dashed line. Adjust the sliders or inputs below the graph to see how changing parameters shifts the shape of the curve.
Interactive PDF Graph
Area Under a Probability Density Function
The relationship between area and probability is the heart of the PDF concept. Every probability statement about a continuous random variable translates into an area problem.
The four facts above work together. The total area is always 1 because probability must sum to certainty. Any specific area between a and b falls between 0 and 1 because it is a fraction of the total. The area at a single point is zero because a point has no width. And the density value f(x) can exceed 1 because it is not an area — it is a height, and multiplying by an appropriately small width still gives an area under 1.
Can a PDF Be Greater Than 1?
Yes. This trips up many students because they expect all probability values to fall between 0 and 1. The key distinction is that f(x) is a density value, not a probability value.
A concrete example: consider the Uniform distribution on the interval [0, 0.5]. The PDF is:
The density is 2 everywhere on the support. Yet the total area is 2 × 0.5 = 1, satisfying the validity condition. The density exceeds 1, but no interval probability does. Density and probability are different objects, and you should never interpret a density value directly as a probability.
Why Is P(X = x) = 0 for a Continuous Random Variable?
For any continuous random variable X and any specific value x, the probability of X equaling exactly that value is zero:
Integration over a degenerate interval — where the lower and upper bounds are the same — always returns zero. Geometrically, a single point has zero width, so the area under the curve above it is zero. This is not a limitation of the theory; it is a feature of what it means to have a continuous probability distribution.
In practice, this means that for a continuous random variable, you cannot ask "what is the probability of exactly 1.75 meters?" The probability of that specific real number is zero. The meaningful question is "what is the probability of falling between 1.74 and 1.76 meters?" That is a range with width, and you can compute it by integrating the PDF over that interval.
A practical consequence: for continuous variables, P(a < X < b) = P(a ≤ X < b) = P(a < X ≤ b) = P(a ≤ X ≤ b). Including or excluding endpoints makes no difference because each endpoint contributes zero probability.
PDF vs CDF
The PDF and the CDF are two ways of describing the same probability distribution. They are related by integration and differentiation.
| Property | PDF — f(x) | CDF — F(x) |
|---|---|---|
| What it gives | Probability density at x | P(X ≤ x) — cumulative probability |
| Can exceed 1? | Yes | No — bounded by [0, 1] |
| Behavior | Can rise and fall | Non-decreasing |
| Limits | Integrates to 1 over all x | F(−∞) = 0, F(∞) = 1 |
| Relationship | f(x) = F′(x) | F(x) = ∫[−∞ to x] f(t) dt |
| Interval probability | P(a≤X≤b) = ∫[a to b] f(x)dx | P(a≤X≤b) = F(b) − F(a) |
| At a point | f(x) measures density | F(x) gives cumulative probability |
The CDF is often more convenient for computing probabilities over intervals because you only need two evaluations: P(a ≤ X ≤ b) = F(b) − F(a). For many distributions — the Normal distribution being the most famous — the PDF has no closed-form antiderivative, so the CDF must be computed numerically or looked up in a table. For distributions with tractable antiderivatives (Uniform, Exponential), the CDF is easy to compute exactly.
PDF vs PMF
The PMF (probability mass function) is the discrete counterpart of the PDF. The distinction is fundamental: a PMF applies to discrete random variables, where probability is concentrated at isolated points; a PDF applies to continuous random variables, where probability is spread across intervals.
| Property | PDF — Continuous | PMF — Discrete |
|---|---|---|
| Variable type | Continuous (real-valued) | Discrete (countable values) |
| Point probability | P(X = x) = 0 for every x | P(X = x) can be positive |
| Total | ∫ f(x) dx = 1 | Σ P(X = x) = 1 |
| Probability over range | Area: ∫[a to b] f(x) dx | Sum: Σ P(X = x) for x in range |
| Can exceed 1? | Yes (density can exceed 1) | No — each probability ≤ 1 |
| Examples | Normal, Uniform, Exponential | Binomial, Poisson, Geometric |
A common way to remember the contrast: for a discrete variable, you can ask "what is the probability of rolling exactly 3?" — the PMF gives that directly as a positive number. For a continuous variable, the equivalent question "what is the probability of measuring exactly 1.732 meters?" always has the answer zero; you must specify an interval instead.
The pages on basic probability and on the binomial distribution — which uses a PMF — illustrate the discrete side of this comparison at Statistics Fundamentals.
How to Find Probability From a PDF
The procedure follows a consistent sequence regardless of which distribution you are working with.
Identify the PDF
Write down f(x) explicitly. For standard distributions, look up the formula. For custom PDFs, the function should be given in the problem.
Identify the Interval and Support
Determine the interval [a, b] over which you want probability. Check the support of the distribution — the range where f(x) > 0. If your interval extends outside the support, the PDF is zero there, and those portions of the integral contribute nothing.
Set Up the Integral
Write P(a ≤ X ≤ b) = ∫[a to b] f(x) dx. Adjust the bounds to stay within the support if necessary.
Integrate
Compute the antiderivative of f(x) if it exists in closed form. For the Normal distribution, integration requires numerical methods or the standard normal table because the antiderivative has no closed form.
Evaluate at the Bounds and Interpret
Apply the Fundamental Theorem of Calculus: [F(b) − F(a)]. The result should be a number between 0 and 1. State the probability clearly: "There is a [result]% probability that X falls between a and b."
Worked Examples
Example 1 — Checking Whether a Function Is a Valid PDF
Problem: Is f(x) = 3x² valid for 0 ≤ x ≤ 1 (and 0 otherwise)?
Nonnegativity check: On [0,1], x² ≥ 0, so 3x² ≥ 0 for all x in the support. ✓
Integral check: ∫[0 to 1] 3x² dx = [x³]₀¹ = 1³ − 0³ = 1 ✓
✅ Result: f(x) = 3x² on [0,1] is a valid PDF. Both conditions are satisfied.
Example 2 — A Function That Is Not a Valid PDF
Problem: Is f(x) = x valid for 0 ≤ x ≤ 1?
Nonnegativity: f(x) = x ≥ 0 on [0,1]. ✓
Integral check: ∫[0 to 1] x dx = [x²/2]₀¹ = 0.5 ≠ 1 ✗
✗ Result: f(x) = x on [0,1] fails the second condition — it integrates to 0.5, not 1. To fix it, use f(x) = 2x, which integrates to 1.
Example 3 — Probability From a Uniform Distribution
Problem: A bus arrives at a uniformly random time between 9:00 and 9:20. What is the probability it arrives between 9:05 and 9:12?
Identify f(x): X ~ Uniform[0, 20] (measuring minutes after 9:00). f(x) = 1/20 = 0.05 for 0 ≤ x ≤ 20.
Set up integral: P(5 ≤ X ≤ 12) = ∫[5 to 12] (1/20) dx
Integrate: = (1/20) × [x]₅¹² = (1/20) × (12 − 5) = 7/20 = 0.35
✅ There is a 35% probability the bus arrives between 9:05 and 9:12. For a uniform distribution, this equals (interval width)/(total width) = 7/20.
Example 4 — Probability From an Exponential Distribution
Problem: Customer service calls arrive at rate λ = 0.5 per minute. What is the probability the wait time between calls is between 2 and 4 minutes?
Identify f(x): X ~ Exponential(λ = 0.5). f(x) = 0.5e^(−0.5x) for x ≥ 0.
Set up integral: P(2 ≤ X ≤ 4) = ∫[2 to 4] 0.5e^(−0.5x) dx
Integrate: Antiderivative of λe^(−λx) is −e^(−λx).
= [−e^(−0.5x)]₂⁴ = −e^(−2) − (−e^(−1)) = e^(−1) − e^(−2)
= 0.3679 − 0.1353 = 0.2326
✅ There is approximately a 23.3% probability the wait time falls between 2 and 4 minutes.
Example 5 — Finding Expected Value From a PDF
Problem: Find E[X] for the PDF f(x) = 3x² on [0, 1].
Formula: E[X] = ∫[−∞ to ∞] x · f(x) dx = ∫[0 to 1] x · 3x² dx
Simplify integrand: x · 3x² = 3x³
Integrate: ∫[0 to 1] 3x³ dx = [3x⁴/4]₀¹ = 3/4 = 0.75
✅ E[X] = 0.75. The expected (average) value of X is 0.75, which lies in the upper portion of [0,1] because the density f(x) = 3x² is higher near x = 1.
Example 6 — Finding Variance From a PDF
Problem: Find Var(X) for f(x) = 3x² on [0, 1], given E[X] = 0.75.
Find E[X²]: E[X²] = ∫[0 to 1] x² · 3x² dx = ∫[0 to 1] 3x⁴ dx = [3x⁵/5]₀¹ = 3/5 = 0.6
Apply the formula: Var(X) = E[X²] − (E[X])² = 0.6 − (0.75)² = 0.6 − 0.5625 = 0.0375
Standard deviation: σ = √0.0375 ≈ 0.194
✅ Var(X) = 0.0375 and σ ≈ 0.194. The distribution is concentrated relatively close to the mean of 0.75.
Example 7 — Finding an Unknown Constant
Problem: f(x) = cx² for 0 ≤ x ≤ 2. Find the constant c that makes this a valid PDF.
Apply the total integral condition: ∫[0 to 2] cx² dx = 1
Integrate: c · [x³/3]₀² = c · (8/3 − 0) = 8c/3
Solve for c: 8c/3 = 1 → c = 3/8 = 0.375
Verify nonnegativity: f(x) = (3/8)x² ≥ 0 on [0, 2] ✓
✅ c = 3/8. The valid PDF is f(x) = (3/8)x² for 0 ≤ x ≤ 2.
Example 8 — Finding a CDF From a PDF
Problem: For f(x) = 2x on [0, 1], find F(x) = P(X ≤ x).
Apply the definition: F(x) = ∫[0 to x] 2t dt for 0 ≤ x ≤ 1
Integrate: F(x) = [t²]₀ˣ = x²
Complete the CDF: F(x) = 0 for x < 0; F(x) = x² for 0 ≤ x ≤ 1; F(x) = 1 for x > 1
Verify: F′(x) = 2x = f(x) ✓. At x=1: F(1) = 1² = 1 ✓
✅ F(x) = x² for 0 ≤ x ≤ 1. You can now find any interval probability: e.g., P(0.3 ≤ X ≤ 0.7) = F(0.7) − F(0.3) = 0.49 − 0.09 = 0.40.
Example 9 — Probability for the Normal Distribution
Problem: X ~ Normal(μ = 100, σ = 15) represents IQ scores. What fraction of people score between 85 and 115?
Standardize: Z = (X − μ)/σ. For X = 85: Z = (85 − 100)/15 = −1. For X = 115: Z = (115 − 100)/15 = 1.
State the probability: P(85 ≤ X ≤ 115) = P(−1 ≤ Z ≤ 1). This is the Empirical Rule interval.
Use the standard normal CDF: P(−1 ≤ Z ≤ 1) = Φ(1) − Φ(−1) = 0.8413 − 0.1587 ≈ 0.6827. See the z-table for Φ values.
✅ Approximately 68.27% of people score between 85 and 115. This reflects the Empirical Rule: about 68% of normally distributed data falls within one standard deviation of the mean. Explore this further in the Empirical Rule guide.
Example 10 — Real-World Application: Reliability Engineering
Problem: A light bulb's lifetime X (in hundreds of hours) follows an Exponential distribution with λ = 0.1. What is the probability a bulb lasts more than 15 hundred hours?
f(x) = 0.1e^(−0.1x) for x ≥ 0. The question asks for P(X > 15) = 1 − F(15).
CDF of Exponential: F(x) = 1 − e^(−λx). So F(15) = 1 − e^(−0.1 × 15) = 1 − e^(−1.5).
Compute: e^(−1.5) ≈ 0.2231. So F(15) ≈ 0.7769 and P(X > 15) = 1 − 0.7769 = 0.2231.
✅ There is approximately a 22.3% probability the bulb lasts more than 1,500 hours. This also illustrates the memoryless property: P(X > s + t | X > s) = P(X > t) for the Exponential distribution.
Expected Value and Variance From a PDF
The expected value (mean) and variance of a continuous random variable are both computed by integrating against the PDF. These are the continuous analogues of the weighted sums used for discrete variables.
x = possible value of X
f(x) = density at x
The product x·f(x) weights each value by its density
E[X²] = ∫ x² · f(x) dx
μ = E[X]
The shortcut formula Var(X) = E[X²] − μ² is almost always more convenient than integrating (x − μ)² · f(x) directly. Compute E[X] and E[X²] separately, then subtract the square of the mean. The standard deviation σ = √Var(X) is in the same units as X, making it easier to interpret than the variance.
Common Probability Density Functions
Each continuous distribution has its own PDF formula, determined by its parameters. The following subsections give the PDF, its support, and the key parameters for the most frequently used continuous distributions.
Normal Distribution
The Bell Curve — Most Widely Used in Statistics
The normal distribution is defined by its mean μ and standard deviation σ. The curve is bell-shaped, symmetric about μ, and extends over all real numbers. The standard normal distribution is the special case where μ = 0 and σ = 1, written Z ~ N(0,1). Any normal variable can be standardized using Z = (X − μ)/σ, making the z-table applicable to all normal distributions.
A complete treatment of the normal distribution, including the Empirical Rule and standardization, is available at Statistics Fundamentals. The z-score page explains the standardization transformation in detail.
Uniform Distribution
Constant Density Over a Bounded Interval
Every point in [a, b] is equally likely in terms of density. This is the continuous distribution of maximum ignorance: if you know a value lies in an interval but have no other information, the Uniform distribution is the natural choice. Because the density is constant, probability over any sub-interval is simply (sub-interval width)/(total width).
Exponential Distribution
Time Between Events in a Poisson Process
The exponential distribution models waiting times and lifetimes. λ > 0 is the rate parameter (events per unit time); the mean waiting time is 1/λ. Its defining property is memorylessness: P(X > s + t | X > s) = P(X > t). This means past waiting time gives no information about future waiting time. Related to the discrete Poisson distribution, which counts events in a fixed interval.
Other Important Continuous Distributions
| Distribution | PDF Form | Support | Key Application |
|---|---|---|---|
| Gamma | f(x) = (β^α / Γ(α)) x^(α−1) e^(−βx) | x > 0 | Sum of exponentials, queuing |
| Beta | f(x) = x^(α−1)(1−x)^(β−1) / B(α,β) | 0 ≤ x ≤ 1 | Proportions, probabilities |
| Lognormal | f(x) = exp(−(ln x − μ)²/(2σ²)) / (xσ√(2π)) | x > 0 | Income, stock prices |
| Weibull | f(x) = (k/λ)(x/λ)^(k−1) e^(−(x/λ)^k) | x ≥ 0 | Failure times, reliability |
| Student's t | Proportional to (1 + x²/ν)^(−(ν+1)/2) | −∞ to ∞ | Small-sample inference |
| Chi-square | f(x) = x^(ν/2−1) e^(−x/2) / (2^(ν/2) Γ(ν/2)) | x > 0 | Goodness-of-fit tests, variance |
Formula Reference Sheet
| Formula | Expression | Notes |
|---|---|---|
| Valid PDF — Condition 1 | f(x) ≥ 0 | Nonnegativity everywhere |
| Valid PDF — Condition 2 | ∫[−∞ to ∞] f(x) dx = 1 | Total probability = 1 |
| Interval Probability | P(a ≤ X ≤ b) = ∫[a to b] f(x) dx | Area under curve |
| Point Probability | P(X = x) = 0 | Continuous variable only |
| CDF Definition | F(x) = ∫[−∞ to x] f(t) dt | Cumulative probability |
| PDF from CDF | f(x) = F′(x) | Where derivative exists |
| Tail Probability | P(X > x) = 1 − F(x) | Complement rule |
| Expected Value | E[X] = ∫[−∞ to ∞] x · f(x) dx | Mean of distribution |
| E[g(X)] | ∫[−∞ to ∞] g(x) · f(x) dx | Expected value of function |
| Variance | Var(X) = ∫(x−μ)² f(x) dx | Spread about the mean |
| Variance Shortcut | Var(X) = E[X²] − (E[X])² | Almost always easier |
| Std Deviation | σ = √Var(X) | Same units as X |
| Normal PDF | (1/σ√(2π)) exp(−(x−μ)²/(2σ²)) | X ~ N(μ, σ²) |
| Uniform PDF | 1/(b−a) on [a, b] | X ~ Uniform(a, b) |
| Exponential PDF | λe^(−λx) for x ≥ 0 | X ~ Exp(λ), mean = 1/λ |
Common Mistakes With Probability Density Functions
| Mistake | Incorrect Thinking | Correct Understanding |
|---|---|---|
| Treating f(x) as probability | f(2) = 0.4 means P(X = 2) = 40% | f(2) = 0.4 is density. P(X = 2) = 0 for continuous X |
| Assuming f(x) ≤ 1 always | The PDF can't exceed 1 because probability is ≤ 1 | f(x) can exceed 1. Only the integral over all x must equal 1 |
| Confusing PDF and CDF | The CDF and PDF give the same information at each point | PDF = density; CDF = cumulative probability. They are related by integration/differentiation |
| Confusing PDF and PMF | The PDF works the same way for any random variable | PDF is for continuous variables (P(X=x)=0). PMF is for discrete variables (P(X=x) > 0) |
| Ignoring the support | Integrating over [−5, 5] for Exponential(1) | Exponential has support x ≥ 0. Adjust lower bound to 0 |
| Confusing variance and std dev | Var(X) and σ are the same thing | σ² = Var(X); σ = √Var(X). Variance is in squared units; SD is in original units |
| Thinking P(a<X<b) ≠ P(a≤X≤b) | Strict vs non-strict inequalities change the probability | For continuous X, both expressions equal ∫[a to b] f(x)dx. Endpoints have zero probability |
Real-World Applications of Probability Density Functions
PDFs appear in any field where measurements produce continuous values. The following examples illustrate the breadth of applications, though the specific numerical parameters would depend on empirical data from each domain.
Physical Measurements
Heights, weights, and temperatures in large populations often follow distributions well-modeled by the Normal PDF. The normal distribution's PDF describes the density of measurement values around a population mean.
Waiting Times
Time between calls at a call center, between equipment failures, or between radioactive decays is frequently modeled by the Exponential PDF. Its memoryless property makes the math tractable.
Reliability Engineering
Component lifetimes in mechanical and electronic systems are modeled using Exponential or Weibull PDFs. Manufacturers use these models to compute failure probabilities and set warranty periods.
Finance and Economics
Lognormal PDFs model asset prices and returns, where values are necessarily positive and distributions are often right-skewed. Risk analysts integrate PDFs to compute value-at-risk measures.
Machine Learning
Probability density functions underlie generative models, density estimation, and Bayesian inference. Gaussian mixture models and kernel density estimation are PDF-based techniques.
Medicine and Biology
Drug concentration in blood, patient measurement distributions, and biological trait distributions all rely on PDF-based modeling for analysis, confidence interval construction, and hypothesis testing.
Frequently Asked Questions
A PDF is a mathematical function that tells you how probability is spread across the values a continuous variable can take. High values of the function mean probability is concentrated there; low values mean it is sparse. You find the actual probability of any range by computing the area under the curve over that range. Probability equals area.
There is no single PDF formula because different distributions have different formulas. The universal formula for computing probability is P(a ≤ X ≤ b) = ∫[a to b] f(x) dx. This says probability equals the integral, or area, of f(x) over the interval [a, b]. Every valid PDF also satisfies f(x) ≥ 0 and ∫[-∞ to ∞] f(x) dx = 1.
No. A PDF value f(x) is a density, not a probability. For a continuous random variable, the probability of any exact value is zero: P(X = x) = 0. Probability comes from integrating f(x) over an interval. The PDF can even exceed 1, which would be impossible for a probability. Thinking of f(x) as a probability is the most common conceptual error with PDFs.
Yes. A PDF value can exceed 1 as long as the total area under the curve equals 1. Example: f(x) = 2 on [0, 0.5] is a valid PDF because ∫[0 to 0.5] 2 dx = 1. Another example: f(x) = 3x² on [0, 1] reaches a maximum of 3 at x = 1. What cannot exceed 1 is any probability computed by integrating over an interval.
The total area under any valid PDF over its entire support equals exactly 1. This follows from the definition: since the variable must take some value, all probabilities together must sum to 1, and probability corresponds to area. The area over any sub-interval [a, b] gives the probability P(a ≤ X ≤ b) and always falls in [0, 1].
The PDF f(x) gives probability density. It describes how probability is concentrated at each point. The CDF F(x) = P(X ≤ x) gives cumulative probability, the probability of being at or below x. They are related by F(x) = ∫[-∞ to x] f(t) dt and f(x) = F′(x). Both contain the same information. The CDF is non-decreasing from 0 to 1; the PDF can take any nonnegative value.
A PMF (probability mass function) applies to discrete random variables and gives the probability of each exact outcome: P(X = x) can be positive. A PDF applies to continuous random variables: P(X = x) = 0 for every exact value, and probability is found by integration over intervals. The PMF sums to 1; the PDF integrates to 1. Examples: the Binomial distribution uses a PMF; the Normal distribution uses a PDF.
The mean (expected value) of a continuous random variable X is E[X] = ∫[-∞ to ∞] x · f(x) dx. You multiply each possible value x by the density f(x) at that value, then integrate over the full support. For standard distributions there are closed-form answers: Normal has mean μ; Uniform on [a,b] has mean (a+b)/2; Exponential with rate λ has mean 1/λ.
Variance is computed using the shortcut formula Var(X) = E[X²] − (E[X])², where E[X²] = ∫[-∞ to ∞] x² · f(x) dx. Compute E[X²] by integrating x² · f(x) over the support, then subtract the square of the mean. The standard deviation is σ = √Var(X).
Exactly two conditions must hold: (1) f(x) ≥ 0 for all x, meaning the function is never negative. (2) ∫[-∞ to ∞] f(x) dx = 1, meaning the total area equals 1. Both conditions must be satisfied simultaneously. A function can be nonnegative but integrate to 0.8, which is not valid, or it can integrate to 1 but dip below zero, which is also not valid. Both conditions are necessary.
The normal distribution PDF is f(x) = (1/(σ√(2π))) · exp(−(x−μ)²/(2σ²)), defined for all real x. It has two parameters: μ, the mean, which determines the center, and σ, the standard deviation, which determines the spread. The standard normal has μ = 0 and σ = 1. The PDF has no closed-form antiderivative, so probabilities are computed using the z-table or numerically.
Because P(X = x) = ∫[x to x] f(t) dt = 0. Integration over a zero-width interval always returns zero, regardless of the density. Geometrically, a single point on the x-axis has no area, so there is no area under the curve above it. This is the natural consequence of working with real-valued continuous measurements, where the probability of hitting any pre-specified exact real number is essentially zero.