Uniform Distribution Formula: Definition, Examples

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Namrata Das

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Uniform distribution is a sort of probability distribution in statistics in which all outcomes are equally probable. A uniform distribution also called a rectangle distribution, is a probability distribution with a constant value. A distribution is a basic graph that depicts a set of data. It can be viewed as either a graph or a list. It reveals which values of a random variable have a lower or higher probability of occurring. There are many distinct forms of probability distributions, with the uniform distribution being the most basic. Discrete and continuous uniform distributions are the two forms of uniform distributions. A discrete uniform distribution produces discrete outcomes with the same probability. A continuous uniform distribution produces outcomes that are both continuous and infinite.

Key Takeaways: Uniform distributions, probability distribution, random variable, discrete and continuous, mean, constant value, statistics

Also read: Frequency Polygon


What is Uniform Distribution?

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A uniform distribution is a type of probability distribution in which every event inside a certain interval has the same chance of happening. It's a graphical representation of a set of information and can be viewed as a graph or a list. In uniform distribution, the random variable is a continuous random variable that is plotted along the x-axis.

If the amplitude of the uniform distribution function remains constant between two points, say a and b, and is 0 otherwise, a continuous random variable is said to follow a uniform distribution. The value of the amplitude is equal to the reciprocal of the length of the range of the continuous random variable in question because the area under the curve of a probability distribution function is always one. The graph of a uniform distribution closely resembles a geometric rectangle.

Example of Uniform Distribution

A Rolling Die: When a fair die is rolled, the probability is that the number on the top of the die is between one and six has a uniform distribution. The probability of number 'one' appearing on top of the die is 1/6, which is the same as the probability of number 'two' appearing on top of the die, and so on. Because each number has an equal chance of appearing at the top, the distribution is uniform.

Coin Tossing: When you flip a coin, the probability of the coin falling with its head up is equal to the probability of the coin landing with its tail up. Because the experiment of tossing the coin has two equally likely outcomes, it is considered to follow a uniform distribution.

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Types of Uniform Distribution

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Uniform distribution can be categorised into two major categories based on the sorts of probable experiment outcomes:

Uniform Discrete Distribution

A discrete uniform distribution is a statistical and probability distribution in which the likelihood of events occurring is equal and falls within a finite range of values. Rolling a six-faced fair die, flipping a coin, and so on are examples of discrete uniform distributions.

In business and management, the concept of discrete uniform distribution is quite useful. It can be used to create a probability distribution that will help a company make the best use of its resources.

Uniform Continuous Distribution

A statistical and probability distribution with an endless number of equally likely values is known as a continuous uniform distribution. Any real value within the defined range can be the result of such a uniform distribution. A random number generator is an example of a continuous uniform distribution.

Also read: Chance and Probability


What is Uniform Distribution Formula?

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The probability density function or probability distribution of a uniform distribution with a continuous random variable X is f(x)=1/b-a, is given by U(a,b), where a and b are constants such that a<x<b.

It is written as follows:

f(x) = 1/ (b-a) for a≤ x ≤b.

where,

a is the smallest possible value.

b is the greatest possible value.

In terms of mean μ and variance σ2, the probability density is stated as follows

Also read: Rolle's Theorem


The Mean μ and Variance σ2 Of Uniform Distribution

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The mean of a uniform distribution variable X is:

E(X) = (1/2) (a + b)

This is also written as:

E(X) = (b + a) / 2.

In the formula, "a" represents the distribution's minimum value, and "b" represents the distribution's maximum value.

The variance of a uniform distribution variable is:

Var(x) = (1/12)(b-a)2

For the above image, the variance is (1/12)(3 – 1)2= 1/12 * 4 = 1/3.


Things to Remember

  • Uniform distribution is a sort of probability distribution in statistics in which all outcomes are equally probable.
  • A uniform distribution also called a rectangle distribution, is a probability distribution with a constant value.
  • Discrete and continuous uniform distributions are the two forms of uniform distributions.
  • It is a simple graphical representation of a set of data. It can be displayed as a graph or as a list.
  • A Rolling Die, Coin Tossing are some of the examples of uniform distributions.
  • Uniform distribution with a continuous random variable X is f(x)=1/b-a, is given by U(a,b), where a and b are constants such that a<x<b.
  • The mean of a uniform distribution variable X is: E(X) = (1/2) (a + b) which is also written as:E(X) = (b + a) / 2.

Also Read:


Sample Questions

Ques: The distillation temperature (T C) is critical in determining the quality of the final product in the production of petroleum. T can be regarded of as a random variable with a uniform distribution ranging from 150°C to 300°C. 1 gallon of petroleum costs £C1 to create. If the oil distils at temperatures below 200 degrees Celsius, the result sells for £C2 per gallon. If it is distilled at a temperature higher than 200 degrees Celsius, it sells for £C3 per gallon. Calculate the projected net profit per gallon. (2 marks)

Ans: P(X < 200) = 50 × 1 150 = 1 3 P(X > 200) = 2 3

Assume F is a random variable that represents profit.

F can take two values £(C2 − C1) or £(C3 − C1) x C2 − C1 C3 − C1 P(F = x) 1/3 2/3 E(F) = C2 − C1 3 + 2 3 [C3 − C1] = C2 − 3C1 + 2C3

Ques: The nominal net weight of the packages is 1 kg. Their actual net weights, on the other hand, have a consistent distribution from 980 g to 1030 g.
(a) Evaluate the probability that a package's net weight is less than 1 kg.
(b) Determine the likelihood that a package's net weight is less than w g, where 980 w 1030.
(c) If package net weights are constant, estimate the probability that all five net weights in a sample of five packages are smaller than wg, and then estimate the probability density function of the largest package's weight. (Hint: if and only if the heaviest package weighs less than w g, each five parcels weigh less than w g.) (2 marks)

Ans: (a) The needed probability is P(W < 1000) = 1000 − 98 1030 − 980 = 20 50 = 0.4

(b) The expected probability is P(W < w) = w − 980 1030 − 980 = w − 980 50

(c) The probability that all five of them are under w g is w 980 50 5, therefore the heaviest pdf is d dw w − 980 50 5 = 5 50 w − 980 50 4 = 0.1 w − 980 50 4 for 980 < w < 1030.

Ques: A person's typical weight increase throughout the winter months is evenly divided between 0 and 30 pounds. Determine the probability that an individual will gain 10 to 15 pounds over the winter months. (2 marks)

Ans: Step 1: Determine the distribution's height.

The area under the curve of a probability distribution is always 1. Because there are 30 units, the height is 1/30. (from 0 to 30).

Step 2: Measure the size of the "slice" of the probability of the question.

Subtract the largest number (b) from the smallest number (a) to get b – a = 15 – 10 = 5.

Step 3: Multiply the width (Step 2) by the height (Step 1) to get the following:

possibility = 5 * 1/30 = 5/30 = 1/6.

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CBSE CLASS XII Related Questions

  • 1.
    Find:

    If \[ (3\hat{i}-2\hat{j}+5\hat{k})\times(4\hat{i}+p\hat{j}+q\hat{k})=\vec{0} \] then find the values of \(p\) and \(q\).

      • \(p = -\frac{2}{3}, \, q = \frac{5}{3}\)
      • \(p = -\frac{8}{3}, \, q = \frac{20}{3}\)
      • \(p = \frac{20}{3}, \, q = -\frac{8}{3}\)
      • \(p = 0, \, q = 0\)

    • 2.
      Find the vector and cartesian equations of the line passing through the point of intersection of the lines \( \vec{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3\hat{i} - \hat{j}) \) and \( \vec{r} = (4\hat{i} - \hat{k}) + \mu(2\hat{i} + 3\hat{k}) \) and parallel to the line \( \frac{x - 1}{-2} = \frac{7 - y}{-3} = z \).


        • 3.
          Find:

          The principal value of \[ \sec^{-1}(\sqrt{2})+2\csc^{-1}(-2) \] is:

            • \(-\frac{\pi}{2}\)
            • \(-\frac{\pi}{4}\)
            • \(\frac{\pi}{4}\)
            • \(\frac{\pi}{2}\)

          • 4.

            A carpenter needs to design a wooden box in the shape of a cuboid such that the sum of its length and breadth is 3 cm more than its height. Twice of its length, thrice of its breadth and its height add up to 10 cm. Its breadth added to 7 times its height is 1 cm less than 3 times its length. 


              • 5.
                Find a point on the line \( \frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2} \) at a distance of \( \sqrt{2} \) units from the point \( (1, 2, 3) \).


                  • 6.

                    Evaluate:
                    \[ \int_{0}^{1} \frac{x \tan^{-1}x}{(1+x^2)^{3/2}}\,dx \]

                      CBSE CLASS XII Previous Year Papers

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