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Edexcel IAL - Statistics 2- 2.2 Probability Density and Distribution Functions- Study notes  - New syllabus

Edexcel IAL – Statistics 2- 2.2 Probability Density and Distribution Functions -Study notes- New syllabus

Edexcel IAL – Statistics 2- 2.2 Probability Density and Distribution Functions -Study notes -Edexcel A level Maths- per latest Syllabus.

Key Concepts:

  • 2.2 Probability Density and Distribution Functions

Edexcel IAL Maths-Study Notes- All Topics

The Probability Density Function and the Cumulative Distribution Function

For a continuous random variable, probabilities are described using a probability density function (pdf) and a cumulative distribution function (cdf).

Probability Density Function (pdf)

A probability density function, denoted by \( f(x) \), satisfies the following conditions:

\( f(x) \geq 0 \) for all \( x \)

The total area under the graph of \( f(x) \) is 1

\( \displaystyle \int_{-\infty}^{\infty} f(x)\,\mathrm{d}x = 1 \)

The function \( f(x) \) itself does not give probabilities. Probabilities are obtained by finding areas under the curve.

Using the Probability Density Function

For a continuous random variable \( X \) with pdf \( f(x) \), the probability that \( X \) lies in an interval is given by

\( \mathrm{P}(a < X \leq b) = \displaystyle \int_{a}^{b} f(x)\,\mathrm{d}x \)

Since \( \mathrm{P}(X=a)=0 \), the probabilities

  • \( \mathrm{P}(a < X \leq b) \)
  • \( \mathrm{P}(a \leq X < b) \)
  • \( \mathrm{P}(a \leq X \leq b) \)

are all equal.

Cumulative Distribution Function (cdf) 

The cumulative distribution function of a continuous random variable \( X \) is defined by

\( \mathrm{F}(x_0) = \mathrm{P}(X \leq x_0) = \displaystyle \int_{-\infty}^{x_0} f(x)\,\mathrm{d}x \)

The cdf gives the probability that the random variable takes a value less than or equal to a given number.

Relationship Between pdf and cdf

For a continuous random variable:

  • The cdf is obtained by integrating the pdf
  • The pdf is the derivative of the cdf, where it exists
  • \( \mathrm{f}(x) = \dfrac{\mathrm{d}}{\mathrm{d}x}\,\mathrm{F}(x) \)

Form of the Probability Density Function

In this syllabus, probability density functions will be restricted to:

  • Simple polynomial functions
  • Piecewise-defined functions

These forms allow probabilities and cumulative distribution functions to be found using straightforward integration.

Example :

A continuous random variable \( X \) has probability density function

\( f(x) = 3x^2,\; 0 \leq x \leq 1 \)

Find \( \mathrm{P}(0.2 < X \leq 0.5) \).

▶️ Answer/Explanation

\( \mathrm{P}(0.2 < X \leq 0.5) = \displaystyle \int_{0.2}^{0.5} 3x^2\,\mathrm{d}x \)

\( = \left[ x^3 \right]_{0.2}^{0.5} = 0.125 – 0.008 = 0.117 \)

Conclusion: The required probability is 0.117.

Example :

A continuous random variable \( X \) has probability density function

\( f(x) = kx,\; 0 \leq x \leq 2 \)

Find the value of \( k \) and hence determine the cumulative distribution function \( \mathrm{F}(x) \) for \( 0 \leq x \leq 2 \).

▶️ Answer/Explanation

Since the total area is 1:

\( \displaystyle \int_{0}^{2} kx\,\mathrm{d}x = 1 \)

\( k\left[\dfrac{x^2}{2}\right]_{0}^{2} = 1 \Rightarrow 2k = 1 \Rightarrow k = \dfrac{1}{2} \)

Now find the cdf:

\( \mathrm{F}(x) = \displaystyle \int_{0}^{x} \dfrac{1}{2}t\,\mathrm{d}t = \dfrac{x^2}{4},\; 0 \leq x \leq 2 \)

Conclusion: \( k = \dfrac{1}{2} \) and \( \mathrm{F}(x) = \dfrac{x^2}{4} \).

Example :

A continuous random variable \( X \) has cumulative distribution function

\( \mathrm{F}(x) = \begin{cases} 0, & x < 0 \\ x^2, & 0 \leq x \leq 1 \\ 1, & x > 1 \end{cases} \)

Find the probability density function \( f(x) \) and determine \( \mathrm{P}(X \leq 0.6) \).

▶️ Answer/Explanation

Differentiate the cdf:

\( f(x) = \dfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{F}(x) = 2x,\; 0 \leq x \leq 1 \)

Now find the probability:

\( \mathrm{P}(X \leq 0.6) = \mathrm{F}(0.6) = (0.6)^2 = 0.36 \)

Conclusion: \( f(x) = 2x \) for \( 0 \leq x \leq 1 \) and the required probability is 0.36.

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