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CIE A-Level Maths Study Notes

5.3.1 Fundamentals of Continuous Random Variables

This section focuses on continuous random variables, a fundamental concept in advanced mathematics. Distinct from discrete variables, continuous random variables can take on an endless array of values, which is crucial in areas like statistics and engineering. The focus will be on understanding continuous random variables, exploring the characteristics of probability density functions (PDFs), and defining the criteria for a valid PDF.

Continuous vs Discrete Random Variables

  • Discrete Random Variables: Have specific, countable outcomes. E.g., the number of pages in a book.
  • Continuous Random Variables: Can take any value within a range. E.g., a person's height.
  • Key Difference: Discrete variables are countable; continuous variables span an infinite range within intervals.

Probability Density Functions (PDFs)

  • PDF: Describes how likely a continuous variable is to take a certain value. Represented as .
  • Characteristics of a PDF:
    • Non-Negativity: f(x)0f(x) \geq 0 for all xx.
    • Normalization: The total area under the PDF (integral) equals 1: f(x)dx=1\int_{-\infty}^{\infty} f(x) dx = 1.

Valid PDF Conditions

  • Non-Negative: Must be 0\geq 0 everywhere.
  • Total Probability Equals 1: Integral over its domain equals 1.
  • Real-Valued: Outputs real numbers.

Example: Validating a PDF

Example 1: Is f(x)=kxf(x) = kx a Valid PDF for xx in [0,2][0, 2]?

Solution:

  • Non-Negativity: True for xx in [0,2][0, 2].
  • Total Probability Equals 1: The integral of f(x)f(x) over its entire range must equal 1.
  • Finding kk for Total Probability:

02kxdx=1\int_0^2 kx \, dx = 1

02kxdx=12kx202\int_0^2 kx \, dx = \frac{1}{2} kx^2 \Big|_0^2

Evaluating this from 0 to 2:

12k(22)12k(02)=1\frac{1}{2} k(2^2) - \frac{1}{2} k(0^2) = 1

2k=1    k=122k = 1 \implies k = \frac{1}{2}

Thus, the function f(x)=12xf(x) = \frac{1}{2}x for xx in [0,2][0, 2] is a valid PDF.

Example 2: Probability Calculation

Function: f(x)=12x)for(x)in([0,2]f(x) = \frac{1}{2}x ) for ( x ) in ([0, 2]

Question: Calculate the probability that xx falls between 1 and 2.

Solution:

The probability P(1x2)P(1 \leq x \leq 2) is given by the integral of ( f(x) ) from 1 to 2:

P(1x2)=1212xdxP(1 \leq x \leq 2) = \int_1^2 \frac{1}{2}x \, dx

1212xdx=14x212\frac{1}{2} \int_1^2 x \, dx = \frac{1}{4}x^2 \Big|_1^2

Evaluating from 1 to 2:

14(22)14(12)=14(4)14(1)=10.25=0.75\frac{1}{4}(2^2) - \frac{1}{4}(1^2) = \frac{1}{4}(4) - \frac{1}{4}(1) = 1 - 0.25 = 0.75

Therefore, the probability that xx falls between 1 and 2 for our PDF is 0.75.

Probability Calculation Graph


Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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