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# Properties of Continuous Uniform Distribution

The continuous uniform distribution is such that the random variable $$X$$ takes values between $$a$$ (lower limit) and $$b$$ (upper limit). In the field of statistics, $$a$$ and $$b$$ are known as the parameters of the continuous uniform distribution. We cannot have an outcome of either less than $$a$$ or greater than $$b$$.

The probability density function for this type of distribution is:

$${ f }_{ x }\left( x \right) =\frac { 1 }{ b -a } \quad \quad a< x < b$$

$$X \sim U (a,b)$$ is the most commonly used shorthand notation read as “the random variable $$x$$ has a continuous uniform distribution with parameters $$α$$ and $$β$$.”

The total probability is spread uniformly between the two limits. Intervals of the same length have the same probability.

In other words, for all $$a \le x_1 < x_2 \le b$$, we have

$$P(X < a) = P(X > b = 0)$$

And the probability that the random variable will have a value between $$x_1$$ and $$x_2$$ is given as follows:

$$P(x_1 \le X \le x_2) =\cfrac {(x_2 – x_1)}{(b – a)}$$

The mean and variance of continuous uniform distribution is given below:

$$\text{Mean} =\cfrac {(a + b)}{2}$$

$$\text{Variance} =\cfrac {(b – a)^2}{12}$$

#### Example: Probability Density Function

You have been given that $$Y \sim U(100,300)$$.

Calculate $$P(Y > 174)$$ and $$P(100 < Y < 226$$.

Solution

The probability density function is given by:

$$f_x(x) =\cfrac {1}{(300 – 100)} =\cfrac {1}{200}$$

Therefore, each “unit interval” has a probability of $$\frac {1}{200}$$.

Which means that $$P(Y > 174) =\cfrac {(300 – 174)}{200} = \cfrac {126}{200} = 0.63$$.

Similarly, $$P(100 < Y < 226) = 0.63$$ because the interval has the same length as above (126) hence the same probability.

### Cumulative Distribution Function of a Continuous Uniform Distribution

Remember that a cumulative distribution function, $$F(x)$$, gives the probability that the random variable $$X$$ is less than or equal to $$x$$, for every $$x$$ value. It is usually expressed as:

$$F(x)=P(X\leq x)$$

The cumulative distribution function of the continuous uniform distribution increases linearly from $$α$$ to $$β$$.

The CDF is linear over the variable’s range and it is given by:

$$F(x)=P(X\leq x)=\frac{x-a}{b-a}$$

## Question

A random variable $$X$$ is uniformly distributed between 32 and 42. What is the probability that $$X$$ will be between 32 and 40?

1. 8%
2. 10%
3. 80%

Solution

First, you should determine the pdf:

\begin{align*} f_x(x) & =\cfrac {1}{(42 – 32)} \\ &=\cfrac {1}{10} \\ \end{align*}

Therefore,

$$P(32 < Y < 40) =\cfrac {(40 – 32)}{10} = 0.8 \text{ or } 80\%$$

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