interpret the probability of an outcome of a continuous random variable as an area under its density curve, and compute probabilities for uniform and exponential distributions.
Instruction
In Statistics A, random variables took separate values: 0, 1, 2 rainy days. Many quantities are continuous: a waiting time, a temperature, an amount of rain. A continuous random variable can take any value in an interval, so we cannot list its values with probabilities. Instead we use a density curve.
Density curves (PS.4). A density curve lies on or above the horizontal axis and has a total area of exactly 1 underneath it. The probability that X falls between a and b is the area under the curve between a and b. Two consequences follow. First, P(X = a) = 0 for any single value, because a line has no area, so P(X < a) and P(X ≤ a) are equal. Second, the height of the curve is not a probability. It is a density: probability per unit of x.
Think of a density curve as the limit of a relative-frequency histogram. As you collect more data and narrow the bins, a histogram scaled so its total area is 1 settles toward a smooth curve.
The uniform distribution. A bus comes every 12 minutes, and you arrive at a random moment. Your wait W is equally likely to be anywhere from 0 to 12 minutes. Its density is a rectangle of width 12. For the area to be 1, the height must be 1/12. Then:
- P(W > 9) = (12 − 9)(1/12) = 3/12 = 0.25
- P(3 < W < 5) = 2/12 ≈ 0.167
- The mean is the balance point, 6 minutes.
The exponential distribution (PS.7). Some waiting times are better described by a curve that starts high and decays: the time until the next call to a help line, or the next customer at a counter, when events happen at random at a steady average rate. The exponential density with mean μ is
f(x) = (1/μ) e−x/μ, x ≥ 0, so P(X > t) = e−t/μ
(This area formula comes from calculus. You can verify it numerically by adding thin rectangles in a spreadsheet.) Illustrative example: if calls to a help line arrive at random, on average every 4 minutes, then the wait X for the next call is exponential with μ = 4:
- P(X > 6) = e−6/4 = e−1.5 ≈ 0.223
- P(X < 2) = 1 − e−0.5 ≈ 0.393
- The median m solves e−m/4 = 0.5, so m = 4 ln 2 ≈ 2.77 minutes, less than the mean of 4. The long right tail pulls the mean up, just as with skewed data in Statistics A.
The exponential has a surprising property. If you have already waited 2 minutes, the chance of waiting more than 6 more is still e−1.5. It has no memory. This fits truly random arrivals, but not a bus on a schedule, which is why the bus used the uniform model.
Choosing a model. A density curve is a model of a process, not a fact about it. You choose one from the context (is every value equally likely? do events arrive at random?) and then check it against data. The most important continuous model, the normal distribution, is next.
Formative check
Work through these before moving on. They are not graded — they tell you, and your teacher, whether the standard below has landed yet.
For a continuous random variable X, what is P(X = 5)?
A wait uniformly distributed from 0 to 12 minutes has density height ; P(wait > 9) = .
For an exponential waiting time with mean 4 minutes, the median wait is also 4 minutes.
Sketch the exponential density with mean 4 and shade P(2 < X < 6). Compute it with the formula (e^(−2/4) − e^(−6/4)). Then check it in a spreadsheet by adding the areas of 40 thin rectangles of width 0.1 from 2 to 6 under f(x) = 0.25e^(−x/4). Explain in two sentences why the rectangle sum is close but not exact.
How confident are you that you can read probabilities as areas under a density curve and compute them for uniform and exponential models?
Practice
Work these on paper or in your notebook, then open Check your answer. Aim for all of Fluency and Application; try at least one Challenge.
Printable version: this unit’s practice workbook (PDF)
Fluency
Build speed and accuracy with the core skill.
- A wait W is uniformly distributed from 0 to 20 minutes. Find the height of the density, P(W > 15), P(5 < W < 8) and the mean wait.
Check your answer
Answer: Height 1/20; P(W > 15) = 0.25; P(5 < W < 8) = 0.15; mean 10 minutes.Areas of rectangles. - For the same uniform wait, find P(W = 7).
Check your answer
Answer: 0.A single value has no area. - A waiting time is exponential with mean 5 minutes. Find P(X > 5).
Check your answer
Answer: e^(−1) ≈ 0.368.P(X > t) = e^(−t/μ). - For the same exponential wait (mean 5), find P(X < 2).
Check your answer
Answer: 1 − e^(−0.4) ≈ 0.330. - Find the median of an exponential distribution with mean 5.
Check your answer
Answer: 5 ln 2 ≈ 3.47 minutes.Solve e^(−m/5) = 0.5. - An exponential wait has mean 10 minutes. Find P(X > 20).
Check your answer
Answer: e^(−2) ≈ 0.135. - Can a density curve have height greater than 1 anywhere? Explain.
Check your answer
Answer: Yes. Height is density, not probability; only the total area must be 1 (for example, uniform on 0 to 0.5 has height 2). - A density curve is symmetric about 12. What is P(X < 12)?
Check your answer
Answer: 0.5.
Application
Use the skill in context. Show your reasoning.
- (Illustrative) Calls to a school’s attendance line arrive at random, on average every 3 minutes. What is the probability of waiting more than 6 minutes for the next call? More than 9 minutes, given that 3 minutes have already passed?
Check your answer
Answer: P(X > 6) = e^(−2) ≈ 0.135. Given 3 minutes have passed, P(X > 9 | X > 3) = P(X > 6) ≈ 0.135, by the memoryless property. - A light-rail train arrives exactly every 15 minutes, and you arrive at a random time. Which model fits your wait, uniform or exponential? Find P(wait > 10).
Check your answer
Answer: Uniform on 0-15 (arrivals are scheduled, not random); P(wait > 10) = 5/15 ≈ 0.333.The exponential fits random arrivals, not scheduled ones. - For an exponential wait with mean 4 minutes, find P(2 < X < 6) and explain why the mean exceeds the median.
Check your answer
Answer: e^(−0.5) − e^(−1.5) ≈ 0.383. The distribution is skewed right; the long right tail pulls the mean (4) above the median (≈ 2.77).
Challenge
Stretch problems. Expect to think before you write.
- The density f(x) = cx on 0 ≤ x ≤ 2 (and 0 elsewhere) is a valid density curve for one value of c. Find c, then find P(X < 1).
Check your answer
Answer: The area is a triangle: ½ · 2 · 2c = 2c = 1, so c = 0.5. P(X < 1) = ½ · 1 · 0.5 = 0.25.Area under a straight-line density is a triangle.
Review
Keep earlier skills sharp.
- X ~ Binomial(n = 15, p = 0.3). Find P(X = 2), P(X ≤ 2), and the mean and standard deviation of X.
Check your answer
Answer: P(X = 2) ≈ 0.0916; P(X ≤ 2) ≈ 0.1268; mean 4.50; SD ≈ 1.775P(X = k) = C(15, k) pk(1 − p)n−k. - A game pays −10 with probability 4/13; 5 with probability 5/13; 50 with probability 4/13. Find the expected value.
Check your answer
Answer: 185/13 ≈ 14.23Multiply each value by its probability and add.