###### Dependent and Independent Variables

Linear regression forecasts the value of a dependent variable given the value of... **Read More**

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

The probability density function for this type of distribution is:

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

\(X \sim U (\alpha, \beta)\) 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 (1) is spread **uniformly** between the two limits. **Intervals of the same length have the same probability**.

**Properties of the Continuous Uniform Distribution**

- For all \(\alpha \le x_1 < x_2 \le \beta\)
- \(P(X < \alpha) = P(X > \beta = 0)\)
- \(P(x_1 \le X \le x_2) =\cfrac {(x_2 – x_1)}{(\beta – \alpha)}\)

$$ \text{Mean} =\cfrac {(\alpha + \beta)}{2} $$

$$ \text{Variance} =\cfrac {(\beta – \alpha)^2}{12} $$

You have been given that \(Y \sim U(100,300)\). Calculate \(P(Y > 174)\) and \(P(100 < Y < 226\)

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.

The cumulative distribution function of the continuous uniform distribution looks like this:

The CDF is linear over the variable’s range.

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

- 8%
- 10%
- 80%

SolutionThe correct answer is

C.First, you should determine the pdf:

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

Therefore,

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