Home / Courses / Probability / Unit 7
Probability Unit 7: Normal, Exponential and Uniform Distributions
The uniform, exponential and normal distributions are the core continuous models, with the gamma as a natural extension. This unit covers their formulas, key properties and the normal approximation to the binomial.
Covers: Uniform distribution, Exponential distribution and memorylessness, Normal distribution and standardization, Normal percentiles, Normal approximation to the binomial, Gamma distribution and Poisson processes, Linear combinations of independent normals
Study guide
The uniform distribution on [a, b] has constant density 1/(b − a), so the probability of any subinterval is its length divided by b − a. Its mean is the midpoint (a + b)/2 and its variance is (b − a)²/12. Conditioning a uniform on a subinterval gives another uniform on that subinterval. Uniform(0, 1) variables are also the raw material for simulation: if U is uniform, then −ln(U)/λ is exponential with rate λ, which is why many computer programs generate random values this way.
The exponential distribution with rate λ has density λe^(−λx), survival function S(x) = e^(−λx), mean 1/λ, variance 1/λ² and median ln 2/λ. It models waiting times between events in a Poisson process and the lifetimes of items that do not wear out. Its defining feature is memorylessness: P(X > s + t | X > s) = P(X > t), so a used item is as good as new. That makes deductible problems simple, since the excess over a deductible is again exponential with the same mean. The minimum of independent exponentials is exponential with the rates added, and the sum of n independent exponentials with rate λ is gamma with shape n.
The normal distribution N(μ, σ²) has a symmetric bell-shaped density. Every probability is found by standardizing, Z = (X − μ)/σ, and using the standard normal CDF Φ, with Φ(−z) = 1 − Φ(z). Percentiles work in reverse: x = μ + z_p σ, where z_p is a standard normal percentile, such as 1.645 for the 95th and 2.326 for the 99th. About 68%, 95% and 99.7% of values fall within 1, 2 and 3 standard deviations. Linear combinations of independent normals are normal: means combine linearly and variances combine with squared coefficients, so Var(X − Y) = Var(X) + Var(Y).
When n is large, binomial(n, p) is approximately normal with mean np and variance np(1 − p); a common rule is np ≥ 10 and n(1 − p) ≥ 10. Because the binomial is discrete, apply a continuity correction: P(X ≥ 60) becomes P(Y > 59.5) and P(X ≤ 50) becomes P(Y < 50.5). The gamma distribution links waiting times to Poisson counts: the time of the n-th arrival exceeds t exactly when fewer than n arrivals occur by time t. Recognizing these connections turns many hard integrals into quick calculations.
Key ideas
- Uniform probabilities are proportional to interval length.
- The exponential is memoryless, so excess waiting time has the same distribution.
- The minimum of independent exponentials is exponential with summed rates.
- Every normal probability reduces to Φ by standardizing.
- Independent normals combine into a normal; variances add even when subtracting.
- Use a continuity correction when approximating discrete counts by a normal.
- The n-th arrival time of a Poisson process is gamma and links to Poisson counts.
Formulas and rules
Common mistakes
- Dividing by the variance instead of the standard deviation when standardizing.
- Subtracting variances when finding the variance of a difference.
- Confusing the exponential rate λ with its mean 1/λ.
- Forgetting the continuity correction in a normal approximation to a count.
- Ignoring the condition in a memoryless problem, or applying memorylessness to a non-exponential variable.
Before the test, make sure you can
- Compute uniform probabilities, means and variances.
- Use the exponential survival function and memorylessness.
- Standardize normal values and read probabilities from Φ.
- Find normal percentiles and solve for μ or σ from tail probabilities.
- Combine independent normals and compute probabilities for sums and differences.
- Apply the normal approximation to the binomial with continuity correction.
- Use the minimum and sums of exponentials, including the link to Poisson counts.
- Compute expected exponential insurance payments with deductibles and limits.
Words to know
- Uniform distribution
- Constant density 1/(b − a) on [a, b]; mean (a + b)/2, variance (b − a)²/12.
- Exponential distribution
- Density λe^(−λx) for x ≥ 0; mean 1/λ, variance 1/λ².
- Rate parameter
- λ in the exponential, the average number of events per unit time; the mean wait is 1/λ.
- Memoryless property (continuous)
- P(X > s + t | X > s) = P(X > t); held only by the exponential.
- Normal distribution
- Bell-shaped density with mean μ and variance σ², written N(μ, σ²).
- Standard normal
- N(0, 1), with CDF Φ.
- z-score
- z = (x − μ)/σ, the number of SDs a value lies from the mean.
- Empirical rule
- About 68%, 95% and 99.7% of a normal lies within 1, 2 and 3 SDs.
- Continuity correction
- Adjusting a discrete boundary by 0.5 when using a continuous approximation.
- Gamma distribution
- The sum of n independent exponential(λ) variables; mean n/λ, variance n/λ².
- Minimum of exponentials
- For independent exponentials, min is exponential with rate λ₁ + λ₂ + ···.
- Symmetry of Φ
- Φ(−z) = 1 − Φ(z).
- Inverse transform method
- If U is uniform(0, 1), F⁻¹(U) has CDF F; for example −ln(U)/λ is exponential.
Sample practice questions
30 more questions like these, from level 1 to 5, are in the practice set.
What are the mean and variance of a uniform distribution on [a, b]?
- Mean (a + b)/2, variance (b − a)²/4
- Mean (a + b)/2, variance (b − a)/12
- Mean (a + b)/2, variance (b − a)²/12
- Mean (b − a)/2, variance (b − a)²/12
Show the answer and steps
Answer: C) Mean (a + b)/2, variance (b − a)²/12
- The density 1/(b − a) is symmetric about the midpoint, so the mean is (a + b)/2.
- Integrating gives E[X²] − (E[X])² = (b − a)²/12.
- The variance must scale with the square of the width, so (b − a)/12 cannot be right.
An exponential random variable has density λe^(−λx) for x ≥ 0. What is its mean?
- 1/λ
- λ
- ln 2/λ
- 1/λ²
Show the answer and steps
Answer: A) 1/λ
- E[X] = ∫₀^∞ x λe^(−λx) dx = 1/λ.
- The rate λ is the reciprocal of the mean.
- 1/λ² is the variance and ln 2/λ is the median.
X ~ N(500, 100²). Find P(400 < X < 650) (to four decimals).
- 0.6826
- 0.0919
- 0.7745
- 0.9332
Show the answer and steps
Answer: C) 0.7745
- z-values: (400 − 500)/100 = −1 and (650 − 500)/100 = 1.5.
- Φ(1.5) − Φ(−1) = 0.9332 − 0.1587.
- = 0.7745.