Probability · Lesson 7 of 9

Continuous distributions

Times, weights and incomes don’t come in steps. For quantities like these the bars of Lesson 5 melt into a smooth curve, sums become integrals, and probability becomes area. Everything else you have learned carries over.

← Probability moduleLesson 7 of 9 · about 45 minutes

A question to hold on to

n students start a problem at the same moment. Each one’s completion time is uniformly distributed on [0, 1], independently. What is the expected time at which the last student finishes?

Delhi School of Economics entrance, 2010, 2017 and 2018Solved at the end ↓

The big shift

Probability becomes area

Pick a number uniformly between 0 and 10. What is the chance it is exactly 5, to infinitely many decimal places? There are infinitely many equally likely numbers, so the chance of any single one is 0. Yet the chance it lands below 5 is plainly ½. For a continuous random variable, probability lives on intervals, not points (JNU 2014 asked exactly this pair).

So we describe it with a probability density function f(x): a curve whose area over an interval is the probability of landing there.

Density

P (a \leq X \leq b) = \int_{a}^{b} f (x) d x, f (x) \geq 0, \int_{- \infty}^{\infty} f (x) d x = 1

Because single points have probability 0, it doesn't matter whether the ends are included: ≤ and < give the same answer.

Choose a density below, then drag across the plot or use the sliders to pick an interval. The shaded area is its probability.

f(x) = 2x/25 on [0, 5] (k = 2/25)

P(1.00 ≤ X ≤ 3.00) mean median

From 1.00To 3.00

P(1.00 ≤ X ≤ 3.00) = F(3.00) − F(1.00)0.3600 − 0.0400 = 0.3200

Mean: 10/3Median: 5/√2 ≈ 3.54Variance: 25/18

Drag on the plot to move the nearer end of the interval.

Show the CDF, F(x) = P(X ≤ x)

A density is not a probability

f(x) can be bigger than 1. A uniform on [0, ½] has height 2 everywhere on it. Only areas under f are probabilities, and the total area is 1. So “a density increases steadily from 0 to 1” is wrong (DSE 2007): that describes a CDF.

The most common first step

Finding the constant

Half the density questions give f up to an unknown constant and let the total area of 1 pin it down. Find the area with the constant in it, set it to 1, and solve.

JNU 2019 and 2021 · standard

f(x) = kx on [0, 5] and 0 elsewhere. Find P(1 ≤ X ≤ 3).

Total area 1.
$\int_{0}^{5} k x d x = k \cdot \frac{25}{2} = 1 ⟹ k = \frac{2}{25}$
Area over [1, 3].
$\int_{1}^{3} \frac{2 x}{25} d x = \frac{x ^{2}}{25}_{1}^{3} = \frac{9 - 1}{25} = \frac{8}{25}$

Answer

8/25. That is the starting set-up in the lab above.

Geometry is often quicker than calculus. DSE 2010’s bakery density is a triangle on [0, 10] with peak 5k at x = 5. Area ½ × 10 × 5k = 25k = 1, so k = 1/25.
Exponentials. f(x) = ae^(−x/5) for x > 0 has area 5a, so a = 0.2 (JNU 2017).
Polynomials. δx(1 − x) on (0, 1) has area δ(½ − ⅓) = δ/6, so δ = 6. Then P(X > 0.3) = 1 − (3x² − 2x³) at 0.3 = 1 − 0.216 = 0.784 (JNU 2020).

DSE 2010 · advanced

For the bakery density (x/25 on [0, 5), (10 − x)/25 on [5, 10)), let A be X ≥ 5 and B be 3 ≤ X ≤ 8. Find P(A | B).

Areas from the triangle.
Left of 3: ½ × 3 × 3/25 = 9/50. Right of 8: ½ × 2 × 2/25 = 4/50. So P(B) = 1 − 13/50 = 37/50.
The overlap.
A ∩ B is [5, 8]: P = ½ − 4/50 = 21/50.
Lesson 3's definition.
P(A | B) = (21/50)/(37/50) = 21/37.

Answer

21/37. Since P(A) = ½ ≠ 21/37, A and B are not independent: DSE’s follow-up question.

The running total

The CDF and the median

The CDF is defined exactly as before, F(x) = P(X ≤ x), but now it is a smooth curve: the area under f to the left of x. Integrate f to get F; differentiate F to get f back. Tick “Show the CDF” in the lab to see the two together.

Density and CDF

F (x) = \int_{- \infty}^{x} f (t) d t, f (x) = F^{'} (x), P (a \leq X \leq b) = F (b) - F (a)

CUET 2024: you get the density by differentiating the CDF, not the other way round, and only for a continuous variable.

So a CDF with slope 1 on [0, 1] and flat afterwards belongs to the density that is 1 on [0, 1] and 0 beyond (DSE 2012): the uniform. The median m is where the CDF crosses one half, F(m) = ½, splitting the area in two.

JNU 2015 · foundation

f(x) = 3x² on [0, 1]. Find λ with P(X ≤ λ) = P(X > λ).

That λ is the median.
Equal probability either side means both are ½.
Integrate.
F(x) = x³ on [0, 1].
Solve F(λ) = ½.
λ³ = ½, so λ = (½)^(1/3) ≈ 0.794.

Answer

(½)^(1/3). JNU 2020 did the same for α(1 − x)^(α−1): F(x) = 1 − (1 − x)^α, and the median is 1 − 2^(−1/α).

Same ideas, integrals for sums

Mean and variance by integration

Every formula from Lesson 5 survives, with Σ replaced by ∫ and p(x) by f(x) dx. The shortcut for the variance and the rules for aX + b are unchanged.

Moments of a continuous variable

E (X) = \int x f (x) d x, E (X^{2}) = \int x^{2} f (x) d x, Var (X) = E (X^{2}) - [E (X)]^{2}

E(g(X)) = ∫ g(x) f(x) dx for any function g.

The uniform on [a, b] has height 1/(b − a), mean (a + b)/2 in the middle, and variance (b − a)²/12. The mean depends on where the interval is; the variance only on how long it is (DSE 2015). For [0, 1]: E(X) = ½, E(X²) = ⅓, Var = 1/12 (JNU 2020). For [−1, 2]: mean ½, variance 9/12 = ¾, median ½ (DSE 2007).

JNU 2016 · standard

X is uniform with mean 1 and variance 4/3. Find P(X < 0).

Length from the variance.
(b − a)²/12 = 4/3 gives b − a = 4.
Position from the mean.
Centred on 1, so the interval is [−1, 3].
Area left of 0.
1 unit of the 4: P(X < 0) = ¼.

Answer

¼.

Not every distribution has a mean. f(x) = 1/x² for x > 1 is a perfectly good density (JNU 2011 used it). But E(X) = ∫ x · x⁻² dx = ∫ dx/x, which grows without bound. It still has a median, 2, where 1 − 1/x = ½. That is why “X has a median” is the statement that is always true (DSE 2014). Try the “1/x²” density in the lab.

The continuous geometric

The exponential: waiting times

How long until the next customer arrives, or a machine fails? If the event is equally likely at every moment, regardless of how long you have already waited, the waiting time is exponential. It is the continuous version of the geometric distribution in Lesson 6.

Exponential with rate λ (mean θ = 1/λ)

f (x) = λ e^{- λ x}, P (X > x) = e^{- λ x}, E (X) = \frac{1}{λ}, Var (X) = \frac{1}{λ ^{2}}, median = \frac{ln 2}{λ}

Some papers use the rate λ, others the mean θ. JNU 2017: a Poisson(λ) and an exponential with rate λ have the same mean only when λ = 1.

Mean above median. ln 2 ≈ 0.69, so the median is about 0.69 of the mean. The long right tail drags the mean up (DSE 2021).
Intervals from the tail. For the standard exponential (λ = 1): P(X > 1) = 1/e, and P(1 < X < 2) = e⁻¹ − e⁻² = (1/e)(1 − 1/e) (JNU 2018).
No memory. P(X > s + t | X > s) = e^(−λ(s+t))/e^(−λs) = e^(−λt) = P(X > t). With mean 50, P(X > 70 | X > 60) = e^(−10/50) = e^(−1/5) (JAM 2024): having lasted 60 tells you nothing.

DSE 2017 · advanced

A machine fails at time T with density (1/3)e^(−t/3). Nobody checks it until t = 2, so a failure before 2 is discovered at 2. What is the expected time of discovery?

Discovery time is max(T, 2).
It is 2 plus any time past 2: max(T, 2) = 2 + (T − 2)⁺.
Use the lack of memory.
(T − 2)⁺ is 0 unless T > 2, which has probability e^(−2/3). Given T > 2, the extra time is again exponential with mean 3.
Add.
E = 2 + e^(−2/3) × 3 = 2 + 3e^(−2/3).

Answer

2 + 3e^(−2/3) ≈ 3.54.

The bell curve

The normal distribution

Heights, measurement errors, test scores: whenever a quantity is the sum of many small independent influences, it tends to follow the normal distribution N(μ, σ²). You never integrate its density by hand in an exam; you use its shape and a table. When it is the logarithm of a quantity that is normal, as with incomes, the quantity is lognormal: see the lognormal and Pareto derivations.

Symmetric about μ. Mean, median and mode are all μ. It is not skewed, and its kurtosis is the benchmark (mesokurtic), so “positively skewed” and “leptokurtic” are both false (CUET 2023). All odd central moments are 0 (Hyderabad 2011).
The 68–95–99.7 rule. About 68.3% of the area lies within 1σ of the mean, 95.4% within 2σ, 99.7% within 3σ. So a score within 2σ of the mean has probability of about 95% (Hyderabad 2014), and one more than 1σ above it about 16% (JNU 2015).
Sums of independent normals are normal. Means add, and variances add after scaling by a²: if X ~ N(5, 4) and Y ~ N(3, 2), then 2X + 3Y ~ N(10 + 9, 4 × 4 + 9 × 2) = N(19, 34) (JAM 2024).

The normal density

f (x) = \frac{1}{σ 2 π} e^{- \frac{1}{2} (\frac{x - μ}{σ})^{2}}

The second parameter is the variance: N(46, 400) has σ = 20. A frequent slip.

Mean μ = 0Standard deviation σ = 1

X ~ N(0, 1)

P(−1.00 ≤ X ≤ 1.00) mean median

From −1.00To 1.00

P(−1.00 ≤ X ≤ 1.00) = F(1.00) − F(−1.00)0.8413 − 0.1587 = 0.6827

Mean = median = mode: μVariance: σ²Here: mean 0, median 0, variance 1

Drag on the plot to move the nearer end of the interval.

Show the CDF, F(x) = P(X ≤ x)

Interactive

Standardising: z-scores

Every normal curve is the standard one, N(0, 1), shifted and stretched. Measure distance from the mean in standard deviations and they all become the same curve. That measure is the z-score, and Φ(z) (the area to the left of z under the standard normal) is what tables give.

Standardising

Z = \frac{X - μ}{σ} \sim N (0, 1), P (X \leq x) = Φ (\frac{x - μ}{σ})

Symmetry gives Φ(−z) = 1 − Φ(z), so a table only needs positive z.

Scores are normal with mean 46 and variance 400. P(score < 30)?

Top row: z, the number of standard deviations from the mean. Bottom row: x, in the question’s own units.

Cut-off x = 30 z = −0.80

z = (x − μ)/σ = (30 − 46)/20 = −0.80

P(X < 30) = P(Z < −0.80) = Φ(−0.80)0.2119

JNU 2012 · foundation

Scores are normal with mean 46 and variance 400. Given Φ(0.8) = 0.788, find P(score < 30).

σ is the square root of the variance.
σ = 20.
Standardise.
z = (30 − 46)/20 = −0.8.
Use symmetry.
Φ(−0.8) = 1 − Φ(0.8) = 1 − 0.788 = 0.212.

Answer

0.212. This is the set-up the tool opens with.

JAM 2026 · standard

P(−1 ≤ Z ≤ 1) = 0.683, P(−2 ≤ Z ≤ 2) = 0.954 and P(−3 ≤ Z ≤ 3) = 0.997. If X ~ N(5, 4), find P(1 ≤ X ≤ 7) to three decimal places.

Standardise both ends (σ = 2).
z = (1 − 5)/2 = −2 and z = (7 − 5)/2 = 1.
Split at the mean, using symmetry.
P(−2 ≤ Z ≤ 0) = 0.954/2 = 0.477 and P(0 ≤ Z ≤ 1) = 0.683/2 = 0.3415.
Add.
0.477 + 0.3415 = 0.8185.

Answer

0.8185, which rounds to 0.818 or 0.819; the official key accepts 0.800 to 0.820. Pick “N(5, 4)” in the tool to see it.

Three facts that answer a surprising number of questions without a table:

Φ(1.645) ≈ 0.95 and Φ(1.96) ≈ 0.975. So μ ± 1.645σ holds 90% of the area: a tyre with mean life 350 and variance 64 lasts between 336.84 and 363.16 days with probability 90% (DSE 2008). And ∫ from 1.96 to ∞ of e^(−x²/2) dx is 0.025√(2π) (DSE 2009), because the density carries a factor 1/√(2π).
E(Z²) = Var(Z) = 1, so for X ~ N(0, σ²), E(X²) = σ² (DSE 2010), and for X ~ N(1, 1), E(X²) = 1 + 1 = 2 (JNU 2013).
Events about X² or X³ are events about X. With P(X < −1) = p, P(X² > 1) = 2p and P(Y³ > 1) = P(Y > 1) = p, so for independent standard normals P(X² > 1 and Y³ > 1) = 2p² (DSE 2004).

The CDF method

Transformations

To find the distribution of Y = g(X), don’t try to transform the density directly. Go through the CDF: write P(Y ≤ y) as a statement about X, use X’s distribution, then differentiate if you need the density.

The CDF method

F_{Y} (y) = P (g (X) \leq y) = P (X \leq g^{- 1} (y)) = F_{X} (g^{- 1} (y)) (g increasing)

For increasing g the inequality keeps its direction. For g(x) = x², it turns into −√y ≤ X ≤ √y.

DSE 2020 · advanced

Y is uniform on [0, 2] and X = e^Y. Find the CDF of X at x ∈ [1, e²].

Rewrite the event.
e^Y ≤ x is the same as Y ≤ ln x, since e^y increases.
Use Y's CDF.
P(Y ≤ ln x) = (ln x)/2.

Answer

F_X(x) = ½ ln x. The same steps give arcsin(x)/π + ½ for X = sin θ with θ uniform on [−π/2, π/2] (DSE 2010 and 2018).

Linear maps of a uniform stay uniform. X uniform on [0, 1] and Y = 2X + 5 gives Y uniform on [5, 7], with density ½ (JNU 2020).
Squares need both sides. X uniform on (−4, 4), Y = X²: P(Y < 2) = P(−√2 < X < √2) = 2√2/8 = √2/4 (JNU 2018).
Absolute values fold the curve. Y = |X| for standard normal X has density 2φ(y) for y > 0, the half-normal: √(π/2) f_Y(y) = e^(−y²/2) (JAM 2023).

Solving the opening question

The largest of n draws

Maxima and minima are the classic use of the CDF method, because “the largest is at most y” means “every one is at most y”, and with independence that is a product.

Maximum and minimum of independent draws

P (max \leq y) = F (y)^{n}, P (min > y) = (1 - F (y))^{n}

For uniforms on [0, 1], F(y) = y.

The opening question · DSE 2010, 2017, 2018

X₁, …, Xₙ are independent and uniform on [0, 1], and Y is the largest. Find E(Y).

CDF of the maximum.
P(Y ≤ y) = yⁿ on [0, 1].
Density by differentiating.
f_Y(y) = n yⁿ⁻¹.
Expectation.
$E (Y) = \int_{0}^{1} y \cdot n y^{n - 1} d y = \frac{n}{n + 1}$

Answer

n/(n + 1). On [0, 2], stretch everything by 2: 2n/(n + 1) (DSE 2020).

There is a picture that makes n/(n + 1) obvious. n random points cut [0, 1] into n + 1 pieces, and by symmetry no piece is longer on average than any other, so each averages 1/(n + 1). The largest point sits one piece short of 1. Drop some points and watch the gaps even out.

Points n = 3

Before you drop any: the points are drawn evenly spaced, 4 gaps of 1/4 each. That is what they look like on average. Now drop some random ones.

Pause and predict

X and Y are independent and uniform on [0, 1], and Z = min(X, Y). What is P(Z < 0.5)? (DSE 2011)

Watch out for

The mistakes examiners are fishing for

Reading N(μ, σ²) as N(μ, σ). N(46, 400) has σ = 20, not 400.
Treating f(x) as a probability. Densities can exceed 1; P(X = x) = 0 for every single x.
Forgetting the constant. Normalise first: kx on [0, 5] is not a density until k = 2/25.
Rate or mean? An exponential “with parameter 5” can mean rate 5 (mean 1/5) or mean 5. Check the density as written.
Squaring without both signs. X² ≤ y means −√y ≤ X ≤ √y.

Revise

The lesson on one screen

Density: P(a ≤ X ≤ b) = ∫ f. f ≥ 0 and ∫ f = 1. P(X = x) = 0, and f can exceed 1.
CDF: F(x) = ∫ f up to x; f = F′. Median: F(m) = ½.
Moments: E(X) = ∫ x f, E(X²) = ∫ x² f, Var = E(X²) − [E(X)]². A mean need not exist; a median always does.
Uniform [a, b]: Height 1/(b − a), mean (a + b)/2, variance (b − a)²/12.
Exponential: P(X > x) = e^(−λx). Mean 1/λ, variance 1/λ², median ln 2/λ < mean. No memory.
Normal: Symmetric: mean = median = mode. 68–95–99.7 within 1, 2, 3σ. Sums of independent normals are normal.
z-score: Z = (X − μ)/σ ~ N(0, 1). Φ(−z) = 1 − Φ(z). 1.645 → 90%, 1.96 → 95% (two-sided).
Transformations: Go through the CDF: F_Y(y) = P(g(X) ≤ y). Squares need both signs.
Max and min: P(max ≤ y) = F(y)ⁿ, P(min > y) = (1 − F(y))ⁿ. Max of n uniforms on [0, 1]: mean n/(n + 1).

Practice

Check yourself

0 of 10 correct so far

Compute

Normalise

f(x) = cx² on [0, 3]. Find c.

Area

f(x) = 1/x² for x > 1 (JNU 2011). Find P(1 < X < 2) + P(4 < X < 5).

Uniform

X is uniform on [2, 8]. Find Var(X).

Var(X)

Exponential

A bulb’s life is exponential with mean 1,000 hours. What is the probability it lasts more than 2,000 hours?

z-score

Weights are normal with mean 60 kg and SD 3 kg (CUET 2023). Using Φ(2) = 0.9772, find P(weight < 66 kg).

Maximum

X₁, X₂, X₃ are independent and uniform on [0, 1]. Find P(max > 0.5).

Quick questions

1For an exponential distribution (DSE 2021)

2X is uniform on (0, 10). P(X = 5) and P(|X| < 5) are (JNU 2014)

3A standard normal variable has (CUET 2022)

4The income distribution of most countries is best described as (JNU 2011)

Real questions

Past papers, by level

Foundation

Definitions and one-step problems. Most CUET and UGC NET questions sit here.

20 questionsPractise →

Standard

Two or three steps, up to the hardest JAM, JNU and Hyderabad questions.

23 questionsPractise →

Delhi School level

This topic’s hardest questions, from the DSE entrance, are worked through in Lesson 9: Distributions.

42 questionsLesson 9 →

All 43 past-paper questions for this lesson, easiest first: practise the full set →

These are questions from the TestStop bank. A free account answers 10 a day, each with its worked solution; All-Access opens every set with no daily limit.

← Lesson 6Discrete distributions Lesson 8 →Sums of random variables, and the inequalities

Continuous distributions

Probability becomes area

Finding the constant

The CDF and the median

Mean and variance by integration

The exponential: waiting times

The normal distribution

Standardising: z-scores

Transformations

The largest of n draws

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next

Continuous distributions

Probability becomes area

Finding the constant

The CDF and the median

Mean and variance by integration

The exponential: waiting times

The normal distribution

Standardising: z-scores

Transformations

The largest of n draws

The mistakes examiners are fishing for

The lesson on one screen

Check yourself

Compute

Quick questions

Past papers, by level

Foundation

Standard

Delhi School level

Read next