Session 4 — Continuous Random Variable
Decision Making Statistics — S04
This session introduces continuous random variables, density functions, and the Normal distribution.
1 Introduction
A random variable \(X\) is continuous if it can take any value in an interval (or a union of intervals).
Unlike a discrete random variable:
- \(P(X=x) = 0\) for any single value \(x\),
- probabilities are computed over intervals such as \(P(a \leq X \leq b)\),
- the model is described by a probability density function \(f(x)\).
For a continuous random variable, \(P(X=x)=0\) does not mean that the value \(x\) is impossible. It means that probability is measured by an area under a curve, not by a single point.
2 Continuous random variable
2.1 Probability density function
A probability density function (pdf) \(f(x)\) describes how probability is distributed along the real line.
Its main properties are:
- \(f(x) \geq 0\) for all \(x\),
- \(\int_{-\infty}^{+\infty} f(x)\,dx = 1\),
- \(P(a \leq X \leq b) = \int_a^b f(x)\,dx\).
For a continuous random variable, probability is an area under the density curve:
\[ P(a \leq X \leq b) = \int_a^b f(x)\,dx \]
2.2 Cumulative distribution function
The cumulative distribution function (CDF) is defined by:
\[ F(x) = P(X \leq x) = \int_{-\infty}^{x} f(t)\,dt \]
Key properties:
- \(F\) is non-decreasing,
- \(F(-\infty)=0\),
- \(F(+\infty)=1\).
\[ P(X \geq a) = 1 - F(a) \]
\[ P(a \leq X \leq b) = F(b) - F(a) \]
2.3 Normal distribution
The most important continuous distribution is the Normal distribution:
\[ X \sim \mathcal{N}(\mu, \sigma^2) \]
where:
- \(\mu\) is the mean,
- \(\sigma^2\) is the variance,
- \(\sigma\) is the standard deviation.
\[ f(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} \]
Main properties:
- bell-shaped curve,
- symmetric around \(\mu\),
- \(E(X)=\mu\),
- \(\operatorname{Var}(X)=\sigma^2\).
2.4 Standard normal distribution and standardization
The standard Normal distribution is:
\[ Z \sim \mathcal{N}(0,1) \]
If \(X \sim \mathcal{N}(\mu,\sigma^2)\), then the standardized variable is:
\[ Z = \frac{X-\mu}{\sigma} \sim \mathcal{N}(0,1) \]
Therefore,
\[ P(X \leq a) = P\left(Z \leq \frac{a-\mu}{\sigma}\right) = \Phi\left(\frac{a-\mu}{\sigma}\right) \]
2.5 Important quantiles of \(\mathcal{N}(0,1)\)
| \(\alpha\) | \(z_{\alpha/2}\) (two-tailed) | \(z_\alpha\) (one-tailed) |
|---|---|---|
| 10% | 1.645 | 1.282 |
| 5% | 1.960 | 1.645 |
| 1% | 2.576 | 2.326 |
2.6 Binomial–Normal approximation
When \(n \geq 30\), \(np \geq 5\), and \(n(1-p) \geq 5\),
\[ B(n,p) \approx \mathcal{N}(np,\,np(1-p)) \]
For Normal distribution questions:
- identify \(\mu\) and \(\sigma\),
- standardize with \(Z=\frac{X-\mu}{\sigma}\),
- read the standard normal table carefully,
- check whether the question asks for a left-tail, right-tail, or interval probability.
3 Application exercise
ExerciseThe heights of adult men in France follow a Normal distribution with mean \(\mu = 176\) cm and standard deviation \(\sigma = 7\) cm.
- What proportion of men are taller than 185 cm?
- What proportion of men are between 170 and 183 cm?
- What is the height that 90% of men are below?
SolutionLet \(X\) be the height of an adult man in France. Then:
\[ X \sim \mathcal{N}(176,7^2) \]
1. Probability of being taller than 185 cm
\[ P(X > 185) = 1 - P(X \leq 185) = 1 - \Phi\left(\frac{185-176}{7}\right) \]
\[ P(X > 185) = 1 - \Phi(1.286) \approx 1 - 0.9006 = 0.0994 \]
2. Probability of being between 170 cm and 183 cm
\[ P(170 \leq X \leq 183) = \Phi\left(\frac{183-176}{7}\right) - \Phi\left(\frac{170-176}{7}\right) \]
\[ P(170 \leq X \leq 183) = \Phi(1) - \Phi(-0.857) \approx 0.8413 - 0.1957 = 0.6456 \]
3. Height below which 90% of men fall
We want \(a\) such that \(P(X \leq a)=0.90\). Using \(z_{0.10}=1.282\):
\[ \frac{a-176}{7} = 1.282 \]
\[ a = 176 + 7 \times 1.282 \approx 184.97 \]
So the required height is approximately 184.97 cm.