SciTech-Mathmatics-Probability+Statistics: Distribution : The Uniform Distribution

SciTech-Mathmatics-Probability+Statistics: Distribution :

The Uniform Distribution

BY ZACH BOBBITTPOSTED ON MARCH 2, 2021

The uniform distribution is a probability distribution in which every value between an interval from a to b is equally likely to occur.

If a random variable \(\large X\) follows a uniform distribution,

then the probability that \(\large X\) takes on a value between \(\large x_1\) and \(\large x_2\) can be found by the following formula:

\(\large \begin{array}{ll} \\
&P(x_1 < X < x_2) = \frac{x_2\ –\ x_1}{b\ –\ a} \\
where:& \\
& x_1:\ the\ \bm{lower\ value}\ of\ \bm{interest} \\
& x_2:\ the\ \bm{upper\ value}\ of\ \bm{interest} \\
&\ \ a:\ the\ \bm{minimum\ possible\ value} \\
&\ \ b:\ the\ \bm{maxmum\ possible\ value} \\
\end{array}\)

For example, suppose the weight of dolphins is uniformly distributed between 100 pounds and 150 pounds.

If we randomly select a dolphin at random, we can use the formula above to determine the probability that the chosen dolphin will weigh between 120 and 130 pounds:

P(120 < X < 130) = (130 – 120) / (150 – 100)

P(120 < X < 130) = 10 / 50

P(120 < X < 130) = 0.2

The probability that the chosen dolphin will weigh between 120 and 130 pounds is 0.2.

Properties of the Uniform Distribution

The uniform distribution has the following properties:

\(\large \begin{array}{ll} \\
\bm{Mean}: ( \bm{ a } + \bm{ b }) / 2 \\
\bm{Median}: ( \bm{ a } + \bm{ b } ) / 2 \\
\bm{Standard\ Deviation}: \sqrt{ \frac{(b\ –\ a)^2}{12} } = \frac{(b\ –\ a)}{\sqrt{12}} \\
\bm{Variance}: \frac{(b\ –\ a)^2}{12} \\
\bm{\sqrt{12}}:\ 3.4641 \\
\end{array}\)

For example, suppose the weight of dolphins is uniformly distributed between 100 pounds and 150 pounds.

We could calculate the following properties for this distribution:

\(\large \begin{array}{ll} \\
Mean\ weight: (a\ +\ b) / 2 = (150\ +\ 100) / 2 = 125 \\
Median\ weight: (a + b) / 2 = (150\ +\ 100) / 2 = 125 \\
Standard\ Deviation\ of\ weight: \sqrt{ \frac{(150\ –\ 100)^2}{12} } = 14.43 \\
Variance\ of\ weight: \frac{(b\ –\ a)^2}{12} = 208.33 \\
\end{array}\)

Uniform Distribution Practice Problems

Use the following practice problems to test your knowledge of the uniform distribution.

Question 1: A bus shows up at a bus stop every 20 minutes. If you arrive at the bus stop, what is the probability that the bus will show up in 8 minutes or less?

Solution 1: The minimum amount of time you’d have to wait is 0 minutes and the maximum amount is 20 minutes. The lower value of interest is 0 minutes and the upper value of interest is 8 minutes.

Thus, we’d calculate the probability as:

P(0 < X < 8) = (8-0) / (20-0) = 8/20 = 0.4.

Question 2: The length of an NBA game is uniformly distributed between 120 and 170 minutes. What is the probability that a randomly selected NBA game lasts more than 155 minutes?

Solution 2: The minimum time is 120 minutes and the maximum time is 170 minutes. The lower value of interest is 155 minutes and the upper value of interest is 170 minutes.

Thus, we’d calculate the probability as:

P(155 < X < 170) = (170-155) / (170-120) = 15/50 = 0.3.

Question 3: The weight of a certain species of frog is uniformly distributed between 15 and 25 grams. If you randomly select a frog, what is the probability that the frog weighs between 17 and 19 grams?

Solution 3: The minimum weight is 15 grams and the maximum weight is 25 grams. The lower value of interest is 17 grams and the upper value of interest is 19 grams.

Thus, we’d calculate the probability as:

P(17 < X < 19) = (19-17) / (25-15) = 2/10 = 0.2.

Note: We can use the Uniform Distribution Calculator to check our answers for each of these problems.