B. Probability Distributions · Continuous Distributions and the Normal DistributionProbability as Area: Ranges, Not Single Values
B2.5 B2.1About 65 minutes
Learning goals- I can explain why P(X = a) = 0 for a continuous random variable, and why probabilities are found over ranges.
- I can find probabilities as areas under uniform and triangular density curves.
- I can use complements and differences of areas, and I know that < and ≤ give the same probability for continuous variables.
- I can connect a rounded measurement to an interval probability.
Words to know
P(a < X < b)
The probability that a continuous random variable X falls between a and b; the area under the density curve from a to b.
continuous uniform distribution
A distribution with density f(x) = 1/(b − a) for a ≤ x ≤ b: every subinterval of the same length has the same probability.
cumulative probability
P(X < a), the area to the left of a.
complement
For any event, P(not A) = 1 − P(A). For example, P(X > a) = 1 − P(X < a).
A strange fact about exact values
You arrive at a bus stop at a random time. Buses come every 20 minutes. What is the probability that you wait exactly 7 minutes? Not 7.01, not 6.999, but exactly 7.000000...?
The answer is 0. That sounds odd, because waiting 7 minutes is clearly possible. This lesson explains why, and shows how to find the probabilities that do matter: probabilities over ranges of values.
Probability = area
From the last lesson: a continuous random variable X has a density curve with total area 1, and
P(a < X < b) = the area under the curve between a and b.
For the bus, every waiting time from 0 to 20 minutes is equally likely, so the density is flat: f(x) = 1/20 = 0.05 for 0 ≤ x ≤ 20. This is a continuous uniform distribution. Its graph is a rectangle 20 wide and 0.05 tall, with area 20 × 0.05 = 1.
Why a single value has probability 0
The area above a single point is a line segment, which has no width, so its area is 0. You can also see it by shrinking an interval around 7 minutes: the probability gets smaller and smaller and approaches 0.
Consequences you will use all the time
- For a continuous X, P(X = a) = 0. Probability 0 does not mean impossible; it means a single value is one of infinitely many.
- So P(a < X < b) = P(a ≤ X ≤ b). Using < or ≤ makes no difference.
- This is different from discrete variables: for the number of heads, P(X ≤ 2) and P(X < 2) are not equal, because P(X = 2) = 3/8.
- Complement: P(X > a) = 1 − P(X < a).
- Between: P(a < X < b) = P(X < b) − P(X < a).
Worked exampleExample 1: Bus waiting times
The waiting time X (min) is uniform from 0 to 20. Find (a) P(5 < X < 12), (b) P(X > 15), (c) P(X ≤ 3 or X ≥ 18), (d) P(X = 10), (e) the mean waiting time.
- (a) Width 12 − 5 = 7, height 0.05: area = 7 × 0.05 = 0.35.
- (b) From 15 to 20: 5 × 0.05 = 0.25.
- (c) Two pieces: 3 × 0.05 + 2 × 0.05 = 0.15 + 0.10 = 0.25.
- (d) A single value: 0.
- (e) The rectangle balances in the middle: (0 + 20)/2 = 10 min.
(a) 0.35 (b) 0.25 (c) 0.25 (d) 0 (e) 10 min
Worked exampleExample 2: A triangular density
The time X (min) students take to finish a 10-minute quiz is modelled by f(x) = x/50 for 0 ≤ x ≤ 10. (Many students use most of the time.) Find P(X < 4), P(X > 8) and P(4 < X < 8).
- Check: the triangle has base 10 and height f(10) = 0.2, so its area is (1/2)(10)(0.2) = 1. ✓
- P(X < a) is a smaller triangle with base a and height a/50: area = (1/2)(a)(a/50) = a²/100.
- P(X < 4) = 16/100 = 0.16.
- P(X < 8) = 64/100 = 0.64, so P(X > 8) = 1 − 0.64 = 0.36.
- P(4 < X < 8) = 0.64 − 0.16 = 0.48.
P(X < 4) = 0.16, P(X > 8) = 0.36, P(4 < X < 8) = 0.48.
The triangular density f(x) = x/50. The shaded area from 4 to 8 minutes is 0.64 − 0.16 = 0.48.Worked exampleExample 3: "Exactly 6 minutes" in real life
A teacher records quiz times to the nearest minute. Using the model in Example 2, what is the probability that a student's time is recorded as 6 minutes?
- "Recorded as 6 minutes" means the true time is in 5.5 ≤ X < 6.5 (half a unit each side).
- P(X < 6.5) = 6.5²/100 = 42.25/100 = 0.4225.
- P(X < 5.5) = 5.5²/100 = 30.25/100 = 0.3025.
- Difference: 0.4225 − 0.3025 = 0.12.
0.12. A "single" recorded value is really a range of true values, so it has a positive probability.
Just for funThrow a dart at a number line from 0 to 1. The probability of hitting any one exact point is 0, yet the dart always lands on some point. Probability 0 is not the same as impossible.
Hands-on activityPaper strip uniform distribution
You need: a strip of paper exactly 20 cm long, a ruler, a sharpened pencil, and a helper.
- Mark the strip from 0 to 20 cm. Shade the part from 5 cm to 12 cm.
- Lay the strip face down. Without looking, have your helper poke it with the pencil tip at a "random" point, 40 times (turning the strip between pokes).
- Flip the strip. Count how many pokes landed in the shaded part, and compare with 40 × 0.35 = 14.
- Now try to count pokes at exactly 10.000 cm. What do you notice?
What to look for: around 14 pokes (anywhere from about 9 to 19 is normal) in the shaded part. People are not perfectly random, so pokes may bunch near the middle. No poke will be at exactly 10.000 cm: you can only say whether it is in a small range around 10.
Your turn
1X is uniform from 0 to 40. Find the height of the density.
2For X uniform from 0 to 40, find P(10 < X < 25).
3For X uniform from 0 to 40, find P(X ≥ 30).
4For X uniform from 0 to 40, find P(X = 12) and P(X ≤ 12).
P(X = 12) = 0; P(X ≤ 12) = 12 × 0.025 = 0.3.
5Using f(x) = x/50 on 0 ≤ x ≤ 10, find P(X < 5).
6Using f(x) = x/50, find P(X > 9).
7Explain why P(X ≤ 3) = P(X < 3) for a continuous X, but not for a discrete X such as the number of heads in 3 tosses.
For continuous X, P(X = 3) = 0, so including 3 adds nothing. For discrete X, P(X = 3) = 1/8, so P(X ≤ 3) = 1 but P(X < 3) = 7/8.
More practice: warm-up
1For a continuous X, what is P(X = 5)?
2What represents P(2 < X < 6) on a density graph?
The area under the curve between 2 and 6.
3P(X < 8) = 0.7. Find P(X > 8).
4P(X < 8) = 0.7 and P(X < 3) = 0.2. Find P(3 < X < 8).
5X is uniform from 10 to 30. What is its mean?
More practice: on your own
1A maple syrup bottling machine shuts off at a random time X (s) uniformly between 2.0 and 3.5 s. Find the density height and P(X > 3.2).
Width 1.5, so height = 1/1.5 ≈ 0.667. P(X > 3.2) = 0.3 ÷ 1.5 = 0.2.
2A streetcar arrives every 12 minutes. You arrive at a random time. Find the probability you wait more than 8 minutes, and the probability you wait between 2 and 5 minutes.
Uniform 0 to 12, height 1/12. P(X > 8) = 4/12 = 1/3. P(2 < X < 5) = 3/12 = 1/4.
3With f(x) = x/50 on 0 ≤ x ≤ 10, find the median time m (the value with P(X < m) = 0.5).
m²/100 = 0.5, so m² = 50 and m = √50 ≈ 7.07 min.
4A length X is recorded as 3 cm (nearest cm). If X is uniform from 0 to 10 cm, what is the probability of this record?
Recorded 3 means 2.5 ≤ X < 3.5: area = 1 × 0.1 = 0.1.
5With f(x) = x/50 on 0 ≤ x ≤ 10, find the probability a time is recorded as 9 minutes (nearest minute).
P(8.5 ≤ X < 9.5) = (9.5² − 8.5²)/100 = (90.25 − 72.25)/100 = 0.18.
Challenge
1A density is f(x) = 0.04x for 0 ≤ x ≤ 5 and f(x) = 0.04(10 − x) for 5 < x ≤ 10 (a symmetric triangle). Show the area is 1 and find P(X < 3) and P(3 < X < 7).
Base 10, peak f(5) = 0.2: area = (1/2)(10)(0.2) = 1 ✓. P(X < 3) = (1/2)(3)(0.12) = 0.18. By symmetry P(X > 7) = 0.18, so P(3 < X < 7) = 1 − 0.36 = 0.64.
2For X uniform on [a, b], show that the probability of an interval of length L inside [a, b] is L/(b − a). Use it: an online class starts at a random time between 9:00 and 9:30. What is the chance it starts in the last 6 minutes?
Height 1/(b − a), width L, so area = L/(b − a). Here 6/30 = 0.2.
3Explain, using areas, why a continuous random variable can have P(X = a) = 0 for every a, yet the total probability is 1.
The total probability is the area over a whole interval, which is made of infinitely many points but has positive width. Each single point has width 0 and so area 0. The points of an interval cannot be listed one by one and added like the values of a discrete variable, so zero-width points never add up to area. For continuous variables, probability comes only from intervals.
Quiz
1. X is uniform from 0 to 50. What is P(20 < X < 35)?
2. For a continuous random variable, P(X = 4) equals...
3. For a continuous X, P(X < 10) = 0.85. What is P(X ≥ 10)?
4. With density f(x) = x/50 for 0 ≤ x ≤ 10, what is P(X < 6)?
5. Which statement is true for a discrete variable X (number of heads in 3 tosses) but false for a continuous one?
6. A time is recorded as 7 s to the nearest second. Which interval should you use to find its probability?
Big idea (B2.5): for continuous variables, probabilities are areas over ranges of values; the probability of any single exact value is 0. That is why < and ≤ give the same answer here, unlike with the discrete distributions in the unit Discrete Probability Distributions.
How to help: Ask your teen to sketch the density and shade the region for every question before calculating. If they can draw it, they can almost always find it. Use a real example: "The bus comes every 15 minutes. What is the chance you wait more than 10?" (5/15 = 1/3).
Common difficulties:
- Thinking P(X = a) = 0 means "impossible".
- Using the density's height as a probability.
- Area of a triangle errors: remember the 1/2.
- Forgetting that a rounded measurement is an interval.