Library
Mathematics

Continuous Uniform Distributions

An exploration of continuous probability density functions, focusing on the uniform distribution, its properties, cumulative accumulation, and event probabilities.

Scene 1 of 4
Probability as Area
f(x) = 0.2x = 1
P(X=1)=∫11f(x)dx=0P(X=1) = \int_1^1 f(x)dx = 0
In a continuous distribution, probability is not about individual points, but about the area under the curve. Because a point has no width, the probability of hitting an exact value like one is zero.
Step-by-step solver
1

(a)

Since X is a continuous random variable, the probability of it taking any single exact value is zero.

P(X=1)=0P(X=1) = 0
2

(b)

For a uniform distribution on [a, b], the mean is (a+b)/2.

E(X)=−3+22=−0.5E(X) = \frac{-3 + 2}{2} = -0.5
3

(c)

The variance of a uniform distribution is given by (b-a)^2 / 12.

Var(X)=(2−(−3))212=2512\text{Var}(X) = \frac{(2 - (-3))^2}{12} = \frac{25}{12}
4

(d)

The CDF F(x) is the integral of the PDF from the lower bound -3 to x.

F(x)=x+35 for −3≤x≤2F(x) = \frac{x+3}{5} \text{ for } -3 \le x \le 2
5

(e)

Solve the inequality X^2 > 1.96, which implies X < -1.4 or X > 1.4, then sum the probabilities.

P(X<−1.4)+P(X>1.4)=F(−1.4)+(1−F(1.4))=0.32+0.12=0.44P(X < -1.4) + P(X > 1.4) = F(-1.4) + (1 - F(1.4)) = 0.32 + 0.12 = 0.44

Original question

A-Level statistics question (see attached image). The graph of the probability density function f(x) of the continuous uniform random variable X is shown. The PDF is constant (height 0.2) from x = -3 to x = 2, and zero elsewhere. (a) Write down P(X = 1). (b) Find E(X). (c) Find Var(X). (d) Sketch the cumulative distribution function of X for -3 <= x <= 2, labelling points where it touches or crosses the axes. (e) Find P(X^2 > 1.96).

Follow-up chat

Ask me anything about this lesson — I'll answer using what we just covered.