Sign in to save

Bookmark this page so you can find it later.

Sign in to save

Bookmark this page so you can find it later.

A uniform distribution models outcomes that are evenly spread across a fixed interval. This cheat sheet helps students recognize when every value in an interval is equally likely and how to calculate probabilities from interval lengths. It is useful for probability density functions, cumulative distribution functions, expected value, variance, and graph interpretation.

Students need these tools to connect geometric area with probability in continuous random variables.

For a continuous uniform random variable XU(a,b)X \sim U(a,b), the probability density is constant from aa to bb and zero outside that interval. Probabilities are found using area, so P(cXd)=dcbaP(c \le X \le d)=\frac{d-c}{b-a} when acdba \le c \le d \le b. The center of the distribution is the mean μ=a+b2\mu=\frac{a+b}{2}, and the spread is measured by σ2=(ba)212\sigma^2=\frac{(b-a)^2}{12}.

The cumulative distribution function increases linearly from 00 to 11 across the interval.

Key Facts

  • For a continuous uniform distribution, write XU(a,b)X \sim U(a,b) where aa is the minimum value and bb is the maximum value.
  • The probability density function is f(x)=1baf(x)=\frac{1}{b-a} for axba \le x \le b and f(x)=0f(x)=0 otherwise.
  • The total area under the density curve is 11, so (ba)1ba=1(b-a)\cdot \frac{1}{b-a}=1.
  • For any interval inside the support, P(cXd)=dcbaP(c \le X \le d)=\frac{d-c}{b-a} when acdba \le c \le d \le b.
  • For a continuous distribution, P(X=c)=0P(X=c)=0 for any single exact value cc.
  • The mean and median of XU(a,b)X \sim U(a,b) are both a+b2\frac{a+b}{2}.
  • The variance is σ2=(ba)212\sigma^2=\frac{(b-a)^2}{12} and the standard deviation is σ=ba12\sigma=\frac{b-a}{\sqrt{12}}.
  • The cumulative distribution function is F(x)=0F(x)=0 for x<ax<a, F(x)=xabaF(x)=\frac{x-a}{b-a} for axba \le x \le b, and F(x)=1F(x)=1 for x>bx>b.

Vocabulary

Uniform distribution
A probability distribution where all values in a given interval are equally likely.
Support
The set of values where a random variable can have nonzero probability density, such as axba \le x \le b for XU(a,b)X \sim U(a,b).
Probability density function
A function f(x)f(x) whose area over an interval gives the probability that XX falls in that interval.
Cumulative distribution function
A function F(x)F(x) that gives the probability P(Xx)P(X \le x).
Expected value
The long-run average value of a random variable, equal to E(X)=a+b2E(X)=\frac{a+b}{2} for a uniform distribution.
Variance
A measure of spread around the mean, equal to σ2=(ba)212\sigma^2=\frac{(b-a)^2}{12} for a uniform distribution.

Common Mistakes to Avoid

  • Using the height as the probability, which is wrong because probability is area under the density curve. For XU(a,b)X \sim U(a,b), use interval length times height, not just 1ba\frac{1}{b-a}.
  • Forgetting that P(X=c)=0P(X=c)=0, which is wrong for continuous random variables because a single point has no width and no area.
  • Using the interval endpoints incorrectly, which can give impossible probabilities. Make sure cc and dd are inside [a,b][a,b] before applying P(cXd)=dcbaP(c \le X \le d)=\frac{d-c}{b-a}.
  • Confusing variance and standard deviation, which is wrong because standard deviation is the square root of variance. For XU(a,b)X \sim U(a,b), σ=ba12\sigma=\frac{b-a}{\sqrt{12}}, not (ba)212\frac{(b-a)^2}{12}.
  • Treating the continuous uniform graph like a histogram with separate bars, which is wrong because the density is a smooth constant height across the whole interval.

Practice Questions

  1. 1 Let XU(2,10)X \sim U(2,10). Find P(4X7)P(4 \le X \le 7).
  2. 2 Let XU(0,24)X \sim U(0,24) represent the time in hours when an event occurs during a day. Find P(X>18)P(X>18).
  3. 3 For XU(5,17)X \sim U(5,17), calculate the mean μ\mu, variance σ2\sigma^2, and standard deviation σ\sigma.
  4. 4 Explain why P(X=6)=0P(X=6)=0 but P(5.9X6.1)P(5.9 \le X \le 6.1) can be greater than 00 for a continuous uniform distribution.