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How do you work with a continuous random variable defined by a probability density function?

Continuous random variables, the probability density function and cumulative distribution function, finding probabilities by integration, and the expectation and variance of a continuous variable.

A focused answer to the OCR A-Level Further Mathematics A Statistics option content on continuous random variables, covering the probability density function and the condition that it integrates to one, finding probabilities by integration, the cumulative distribution function and its relationship to the pdf, and the expectation and variance of a continuous variable.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The probability density function
  3. Finding probabilities
  4. The cumulative distribution function
  5. Expectation and variance
  6. Try this

What this dot point is asking

OCR's Statistics option wants you to handle a continuous random variable described by a probability density function (pdf): use the condition that the pdf integrates to one, find probabilities by integrating the pdf over an interval, relate the pdf to the cumulative distribution function (cdf), and compute the expectation and variance by integration.

The probability density function

For a continuous variable, probability is spread out rather than concentrated at points, so it is described by a density. The density must be non-negative everywhere and enclose a total area of one.

Because individual points have zero area, P(X=c)=0\mathrm{P}(X = c) = 0 and it does not matter whether inequalities are strict, so P(a≤X≤b)=P(a<X<b)\mathrm{P}(a \le X \le b) = \mathrm{P}(a < X < b).

Finding probabilities

A probability is the area under the pdf over the relevant interval, found by definite integration. The first step in many questions is to find an unknown constant by imposing that the total area is one.

The cumulative distribution function

The cdf gives the probability of being at or below a value, accumulating area from the left. It is the integral of the pdf, and conversely the pdf is the derivative of the cdf.

Expectation and variance

The mean and variance use the same ideas as the discrete case, with integration of xf(x)x f(x) and x2f(x)x^2 f(x) in place of summation. The intuition carries over directly: where the discrete mean weights each value by its probability and adds, the continuous mean weights each value by its probability density and integrates, and the variance again measures spread about the mean. For a symmetric pdf the mean sits at the axis of symmetry, which is a quick check on your integration. The median, by contrast, is the value mm for which F(m)=12F(m) = \tfrac{1}{2}, found by solving the cdf equation, and need not equal the mean unless the distribution is symmetric.

Continuous random variables reuse the integration skills of further calculus and parallel the discrete distributions, completing the random-variable toolkit.

Try this

Q1. A pdf is f(x)=cxf(x) = cx for 0≤x≤20 \le x \le 2. Find cc. [2 marks]

  • Cue. ∫02cx dx=2c=1\displaystyle\int_0^2 cx\,dx = 2c = 1, so c=12c = \tfrac{1}{2}.

Q2. Given a cdf F(x)=x29F(x) = \tfrac{x^2}{9} for 0≤x≤30 \le x \le 3, find the pdf. [2 marks]

  • Cue. f(x)=F′(x)=2x9f(x) = F'(x) = \tfrac{2x}{9} on 0≤x≤30 \le x \le 3.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20186 marksA continuous random variable XX has probability density function f(x)=kx2f(x) = kx^2 for 0≤x≤30 \le x \le 3 and f(x)=0f(x) = 0 otherwise. Find kk, and P(X>2)\mathrm{P}(X > 2).
Show worked answer →

The pdf integrates to 11 (M1): ∫03kx2 dx=k[x33]03=k⋅9=9k=1\displaystyle\int_0^3 kx^2\,dx = k\left[\dfrac{x^3}{3}\right]_0^3 = k\cdot 9 = 9k = 1, so k=19k = \tfrac{1}{9} (A1, A1).

Probability (M1): P(X>2)=∫2319x2 dx=19[x33]23\mathrm{P}(X > 2) = \displaystyle\int_2^3 \tfrac{1}{9}x^2\,dx = \tfrac{1}{9}\left[\dfrac{x^3}{3}\right]_2^3 (A1).

Evaluate (A1): =19⋅27−83=19⋅193=1927= \tfrac{1}{9}\cdot\dfrac{27 - 8}{3} = \tfrac{1}{9}\cdot\dfrac{19}{3} = \dfrac{19}{27}.

Markers reward the normalisation integral, the value of kk, the probability integral, and the value 1927\tfrac{19}{27}.

OCR 20226 marksA continuous random variable XX has pdf f(x)=12xf(x) = \tfrac{1}{2}x for 0≤x≤20 \le x \le 2 and 00 otherwise. Find E(X)\mathrm{E}(X) and Var(X)\mathrm{Var}(X).
Show worked answer →

Expectation (M1): E(X)=∫02x⋅12x dx=12∫02x2 dx=12[x33]02=12⋅83=43\mathrm{E}(X) = \displaystyle\int_0^2 x\cdot\tfrac{1}{2}x\,dx = \tfrac{1}{2}\displaystyle\int_0^2 x^2\,dx = \tfrac{1}{2}\left[\dfrac{x^3}{3}\right]_0^2 = \tfrac{1}{2}\cdot\dfrac{8}{3} = \dfrac{4}{3} (A1).

For the variance, first E(X2)\mathrm{E}(X^2) (M1): E(X2)=∫02x2⋅12x dx=12[x44]02=12⋅4=2\mathrm{E}(X^2) = \displaystyle\int_0^2 x^2\cdot\tfrac{1}{2}x\,dx = \tfrac{1}{2}\left[\dfrac{x^4}{4}\right]_0^2 = \tfrac{1}{2}\cdot 4 = 2 (A1).

Apply Var(X)=E(X2)−[E(X)]2\mathrm{Var}(X) = \mathrm{E}(X^2) - [\mathrm{E}(X)]^2 (M1): =2−(43)2=2−169=29= 2 - \left(\dfrac{4}{3}\right)^2 = 2 - \dfrac{16}{9} = \dfrac{2}{9} (A1).

Markers reward the expectation integral, the E(X2)\mathrm{E}(X^2) integral, the variance formula, and the value 29\tfrac{2}{9}.

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