Edexcel IAL - Statistics 2- 2.1 Continuous Random Variables- Study notes - New syllabus
Edexcel IAL – Statistics 2- 2.1 Continuous Random Variables -Study notes- New syllabus
Edexcel IAL – Statistics 2- 2.1 Continuous Random Variables -Study notes -Edexcel A level Maths- per latest Syllabus.
Key Concepts:
- 2.1 Continuous Random Variables
The Concept of a Continuous Random Variable
A continuous random variable is a random variable that can take any value within a given interval. Unlike a discrete random variable, its possible values are not countable.
Examples of continuous random variables include time, mass, height, length, temperature, and pressure.
Key Characteristics
- The probability that a continuous random variable takes any exact value is zero
- Probabilities are found over intervals
- Probabilities are represented using areas under a curve
- Hence, for a continuous random variable \( X \):
\( \mathrm{P}(X = a) = 0 \)
Probability Density Function (PDF)
A continuous random variable is described by a probability density function (pdf), denoted by \( f(x) \).
The pdf satisfies the following properties:
\( f(x) \geq 0 \) for all \( x \)
The total area under the curve is 1
\( \displaystyle \int_{-\infty}^{\infty} f(x)\,\mathrm{d}x = 1 \)
Finding Probabilities
For a continuous random variable \( X \) with pdf \( f(x) \):
\( \mathrm{P}(a \leq X \leq b) = \displaystyle \int_{a}^{b} f(x)\,\mathrm{d}x \)
Since \( \mathrm{P}(X=a)=0 \), inequalities such as
\( \mathrm{P}(a \leq X \leq b) \)
\( \mathrm{P}(a < X \leq b) \)
are all equal.
Cumulative Distribution Function (CDF)
The cumulative distribution function of a continuous random variable is defined by
\( \mathrm{F}(x) = \mathrm{P}(X \leq x) = \displaystyle \int_{-\infty}^{x} f(t)\,\mathrm{d}t \)
Example :
Explain why \( \mathrm{P}(X = 4) = 0 \) for a continuous random variable \( X \).
▶️ Answer/Explanation
A continuous random variable can take infinitely many values in any interval.
Probabilities are found using areas over intervals, not at single points.
Conclusion: The probability at any exact value is zero, so \( \mathrm{P}(X=4)=0 \).
Example :
A continuous random variable \( X \) has probability density function
\( f(x) = 2x,\; 0 \leq x \leq 1 \)
Find \( \mathrm{P}(0.2 \leq X \leq 0.6) \).
▶️ Answer/Explanation
\( \mathrm{P}(0.2 \leq X \leq 0.6) = \displaystyle \int_{0.2}^{0.6} 2x\,\mathrm{d}x \)
\( = \left[ x^2 \right]_{0.2}^{0.6} = 0.36 – 0.04 = 0.32 \)
Conclusion: The required probability is 0.32.
Example :
A continuous random variable \( X \) has pdf
\( f(x) = k(1-x),\; 0 \leq x \leq 1 \)
Find the value of \( k \).
▶️ Answer/Explanation
The total area under the pdf must be 1:
\( \displaystyle \int_{0}^{1} k(1-x)\,\mathrm{d}x = 1 \)
\( k\left[x – \dfrac{x^2}{2}\right]_{0}^{1} = k\left(1 – \dfrac{1}{2}\right) = \dfrac{k}{2} \)
\( \dfrac{k}{2} = 1 \Rightarrow k = 2 \)
Conclusion: The value of \( k \) is 2.