Uniform Distribution Example
A random variable X follows a continuous Uniform distribution between 0 and 1, denoted as X ~ Uniform(0,1).
What is the probability P(0.2 < X < 0.5)?
Related Concepts
Hint
Let's use the simple formula for probabilities in a Uniform distribution:
- Identify the main range [a, b]: The problem states X ~ Uniform(0,1). So, 'a' (the start of the range) is 0, and 'b' (the end of the range) is 1.
- Identify the target interval [c, d]: We want to find the probability P(0.2 < X < 0.5). So, 'c' (the start of our target interval) is 0.2, and 'd' (the end of our target interval) is 0.5.
- Apply the formula: The probability is
(d - c) / (b - a). Plug in the values for a, b, c, and d. - Visualize it: Imagine a line segment from 0 to 1. What fraction of this line is covered by the segment from 0.2 to 0.5?
Explanation: Uniform Distribution Probability
Imagine you have a special random number generator. This generator is set up to pick any number between 0 and 1, and every single number in that range has an absolutely equal chance of being chosen. It won't pick a number less than 0, and it won't pick a number greater than 1.
Think of it like throwing a tiny dart at a 1-meter long ruler. If your aim is perfectly random (but you always hit the ruler), any spot on the ruler is equally likely to be hit.
The Question: What's the chance that the number picked is somewhere between 0.2 and 0.5?
Let's think about lengths:
- The total length where our random number can be is from 0 to 1. The length of this entire range is 1 - 0 = 1.
- The length of the specific part we're interested in is from 0.2 to 0.5. The length of this part is 0.5 - 0.2 = 0.3.
Since every number is equally likely, the probability is just like finding what fraction of the total length our target part covers:
Probability = (Length of our target part) / (Total possible length)
Probability = 0.3 / 1
Probability = 0.3
So, there's a 0.3 (or 30%) chance that the random number will fall between 0.2 and 0.5.
We are given that the random variable X follows a continuous Uniform distribution between 0 and 1. This is written as X ~ Uniform(0,1).
In this notation:
- 'a' is the lower bound of the distribution = 0.
- 'b' is the upper bound of the distribution = 1.
The Probability Density Function (PDF) for a general Uniform distribution U(a,b) is: f(x) = 1 / (b - a) for values of x between a and b (i.e., a ≤ x ≤ b). f(x) = 0 for values of x outside this range.
For our specific case, X ~ Uniform(0,1): f(x) = 1 / (1 - 0) = 1 / 1 = 1, for 0 ≤ x ≤ 1. f(x) = 0, otherwise.
We want to find the probability P(0.2 < X < 0.5).
Method 1: Using the Direct Formula for Uniform Probability
For a continuous uniform distribution X ~ Uniform(a,b), the probability that X falls within a sub-interval [c, d] (where a ≤ c ≤ d ≤ b) is given by:
P(c < X < d) = (d - c) / (b - a)
In our problem:
- a = 0 (lower bound of the distribution)
- b = 1 (upper bound of the distribution)
- c = 0.2 (lower bound of our interval of interest)
- d = 0.5 (upper bound of our interval of interest)
Plugging these values into the formula:
P(0.2 < X < 0.5) = (0.5 - 0.2) / (1 - 0) P(0.2 < X < 0.5) = 0.3 / 1 P(0.2 < X < 0.5) = 0.3
Method 2: Using Integration (More Formal Approach)
The probability P(c < X < d) is found by integrating the PDF, f(x), from c to d. P(0.2 < X < 0.5) = ∫0.20.5 f(x) dx
Since our PDF f(x) = 1 for any x between 0 and 1 (and the interval from 0.2 to 0.5 is within this range), the integral becomes: P(0.2 < X < 0.5) = ∫0.20.5 1 dx
The integral of 1 with respect to x is just x. So we evaluate x at the upper and lower limits: P(0.2 < X < 0.5) = [x]from 0.2 to 0.5 P(0.2 < X < 0.5) = (0.5) - (0.2) P(0.2 < X < 0.5) = 0.3
Final Result
Both methods give the same result. The probability that the random variable X (which is uniformly distributed between 0 and 1) falls between 0.2 and 0.5 is:
P(0.2 < X < 0.5) = 0.3
This means there is a 30% chance that X will be in this interval.
Geometric View: The PDF f(x) = 1 for 0 ≤ x ≤ 1 looks like a rectangle.
- The base of this rectangle goes from x=0 to x=1 (length = 1).
- The height of this rectangle is f(x)=1.
- The total area of this rectangle is base × height = 1 × 1 = 1. (The total probability must always be 1).
- The base of this smaller rectangle goes from x=0.2 to x=0.5 (length = 0.5 - 0.2 = 0.3).
- The height is still f(x)=1.
- The area (and thus the probability) is 0.3 × 1 = 0.3.
The shaded area represents P(0.2 < X < 0.5).
Think About It: If a bus arrives at a stop uniformly between 8:00 AM and 8:10 AM (a 10-minute interval), what is the probability you will wait less than 3 minutes if you arrive exactly at 8:00 AM? How does the PDF f(x) change for this new range?