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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

Formulas and rules

Uniform[a, b]: f = 1/(b − a); E = (a + b)/2; Var = (b − a)²/12
Exponential(λ): f = λe^(−λx); S(x) = e^(−λx); E = 1/λ; Var = 1/λ²; median ln 2/λ
Normal: Z = (X − μ)/σ; x_p = μ + z_p σ; z₀.₉₀ = 1.2816, z₀.₉₅ = 1.645, z₀.₉₇₅ = 1.96, z₀.₉₉ = 2.326
aX + bY with X, Y independent normal: mean aμ_X + bμ_Y, variance a²σ_X² + b²σ_Y²
Gamma(n, λ) as a sum of n exponentials: E = n/λ, Var = n/λ²
Normal approximation: binomial ≈ N(np, np(1 − p)) with ±0.5 continuity correction

Common mistakes

Before the test, make sure you can

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.

Question 1 (level 1)

What are the mean and variance of a uniform distribution on [a, b]?

  1. Mean (a + b)/2, variance (b − a)²/4
  2. Mean (a + b)/2, variance (b − a)/12
  3. Mean (a + b)/2, variance (b − a)²/12
  4. 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

  1. The density 1/(b − a) is symmetric about the midpoint, so the mean is (a + b)/2.
  2. Integrating gives E[X²] − (E[X])² = (b − a)²/12.
  3. The variance must scale with the square of the width, so (b − a)/12 cannot be right.
Question 2 (level 1)

An exponential random variable has density λe^(−λx) for x ≥ 0. What is its mean?

  1. 1/λ
  2. λ
  3. ln 2/λ
  4. 1/λ²
Show the answer and steps

Answer: A) 1/λ

  1. E[X] = ∫₀^∞ x λe^(−λx) dx = 1/λ.
  2. The rate λ is the reciprocal of the mean.
  3. 1/λ² is the variance and ln 2/λ is the median.
Question 3 (level 3)

X ~ N(500, 100²). Find P(400 < X < 650) (to four decimals).

  1. 0.6826
  2. 0.0919
  3. 0.7745
  4. 0.9332
Show the answer and steps

Answer: C) 0.7745

  1. z-values: (400 − 500)/100 = −1 and (650 − 500)/100 = 1.5.
  2. Φ(1.5) − Φ(−1) = 0.9332 − 0.1587.
  3. = 0.7745.