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Statistics and Probability - Continuous random variables

Grade 11IB_AA

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A continuous random variable XX is a variable that can take any value within a given range. Unlike discrete variables, the probability of XX taking a specific exact value is zero: P(X=x)=0P(X = x) = 0.

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The Probability Density Function (PDF), denoted by f(x)f(x), describes the relative likelihood of the variable falling within a particular range. For f(x)f(x) to be a valid PDF, it must satisfy: f(x)≥0f(x) \ge 0 for all xx and ∫−∞∞f(x)dx=1\int_{-\infty}^{\infty} f(x) dx = 1.

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The probability that XX lies between aa and bb is the area under the PDF curve between those points: P(a≤X≤b)=∫abf(x)dxP(a \le X \le b) = \int_{a}^{b} f(x) dx.

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The Cumulative Distribution Function (CDF), denoted by F(x)F(x), represents the probability that the variable is less than or equal to xx: F(x)=P(X≤x)=∫−∞xf(t)dtF(x) = P(X \le x) = \int_{-\infty}^{x} f(t) dt.

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The median mm of a continuous random variable is the value such that F(m)=0.5F(m) = 0.5 or ∫−∞mf(x)dx=0.5\int_{-\infty}^{m} f(x) dx = 0.5.

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The expected value E(X)E(X) (also called the mean μ\mu) represents the long-term average value of the variable.

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The variance Var(X)Var(X) measures the spread of the distribution and is calculated using the second moment E(X2)E(X^2).

📐Formulae

∫−∞∞f(x)dx=1\int_{-\infty}^{\infty} f(x) dx = 1

P(a≤X≤b)=∫abf(x)dxP(a \le X \le b) = \int_{a}^{b} f(x) dx

F(x)=∫−∞xf(t)dtF(x) = \int_{-\infty}^{x} f(t) dt

E(X)=∫−∞∞xf(x)dxE(X) = \int_{-\infty}^{\infty} x f(x) dx

Var(X)=E(X2)−[E(X)]2=∫−∞∞x2f(x)dx−[∫−∞∞xf(x)dx]2Var(X) = E(X^2) - [E(X)]^2 = \int_{-\infty}^{\infty} x^2 f(x) dx - \left[ \int_{-\infty}^{\infty} x f(x) dx \right]^2

f(x)=ddxF(x)f(x) = \frac{d}{dx} F(x)

💡Examples

Problem 1:

A continuous random variable XX has a probability density function given by f(x)={kx20≤x≤30otherwisef(x) = \begin{cases} kx^2 & 0 \le x \le 3 \\ 0 & \text{otherwise} \end{cases}. Find the value of the constant kk.

Solution:

To find kk, we use the property that the total area under the PDF must be 11:

∫03kx2dx=1\int_{0}^{3} kx^2 dx = 1

k[x33]03=1k \left[ \frac{x^3}{3} \right]_0^3 = 1

k(273−0)=1k \left( \frac{27}{3} - 0 \right) = 1

9k=1⇒k=199k = 1 \Rightarrow k = \frac{1}{9}

Explanation:

We integrate the PDF over its defined range and set the result to 11 to solve for the unknown coefficient kk.

Problem 2:

For the PDF f(x)=19x2f(x) = \frac{1}{9}x^2 for 0≤x≤30 \le x \le 3, calculate the mean E(X)E(X).

Solution:

The mean is given by E(X)=∫−∞∞xf(x)dxE(X) = \int_{-\infty}^{\infty} x f(x) dx:

E(X)=∫03x(19x2)dxE(X) = \int_{0}^{3} x \left( \frac{1}{9}x^2 \right) dx

E(X)=∫0319x3dxE(X) = \int_{0}^{3} \frac{1}{9}x^3 dx

E(X)=19[x44]03E(X) = \frac{1}{9} \left[ \frac{x^4}{4} \right]_0^3

E(X)=19(814−0)=94=2.25E(X) = \frac{1}{9} \left( \frac{81}{4} - 0 \right) = \frac{9}{4} = 2.25

Explanation:

We multiply the PDF by xx and integrate over the domain [0,3][0, 3] to find the expected value.

Problem 3:

Find the median mm for a continuous random variable with PDF f(x)=2xf(x) = 2x for 0≤x≤10 \le x \le 1.

Solution:

The median mm satisfies ∫0mf(x)dx=0.5\int_{0}^{m} f(x) dx = 0.5:

∫0m2xdx=0.5\int_{0}^{m} 2x dx = 0.5

[x2]0m=0.5[x^2]_0^m = 0.5

m2−0=0.5m^2 - 0 = 0.5

m=0.5≈0.707m = \sqrt{0.5} \approx 0.707

Explanation:

The median is the value that splits the area under the PDF into two equal halves of 0.50.5 each. We integrate from the lower bound to mm and solve for mm.