Uniformly distributed random variable. Uniform continuous distribution in EXCEL. Uniform Distribution Characteristics

In practice, there are random variables about which it is known in advance that they can take on any value within strictly defined boundaries, and within these boundaries all values ​​of the random variable have the same probability (have the same probability density).

For example, when a clock breaks down, the stopped minute hand will show with equal probability (probability density) the time elapsed from the beginning of the given hour until the clock breaks. This time is a random variable that takes values ​​with the same probability density that do not go beyond the boundaries defined by the duration of one hour. Rounding error also belongs to such random variables. Such quantities are said to be uniformly distributed, that is, they have a uniform distribution.

Definition. A continuous random variable X has a uniform distribution on the interval[a, in], if on this segment the probability distribution density of the random variable is constant, i.e. if the differential distribution function f(x) has the following form:

This distribution is sometimes called law of uniform density. About a quantity that has a uniform distribution on a certain segment, we will say that it is distributed uniformly on this segment.

Find the value of the constant c. Since the area bounded by the distribution curve and the axis Oh, equals 1, then

where With=1/(b-a).

Now the function f(x)can be represented as

Let's construct the distribution function F(x ), for which we find the expression F (x ) on the interval [ a , b]:

Graphs of functions f (x) and F (x) look like:

Let's find numerical characteristics.

Using the formula for calculating the mathematical expectation of the NSW, we have:

Thus, the mathematical expectation of a random variable uniformly distributed on the interval [a , b] coincides with the middle of this segment.

Find the variance of a uniformly distributed random variable:

from which it immediately follows that the standard deviation:

Let us now find the probability that the value of a random variable with a uniform distribution falls into the interval(a , b ) , belonging entirely to the segment [a,b ]:

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As mentioned earlier, examples of probability distributions continuous random variable X are:

• uniform probability distribution of a continuous random variable;
• exponential probability distribution of a continuous random variable;
• normal distribution probabilities of a continuous random variable.

Let us give the concept of uniform and exponential distribution laws, probability formulas and numerical characteristics of the considered functions.

Index	Random distribution law	The exponential law of distribution
Definition	Uniform is called the probability distribution of a continuous random variable X, whose density remains constant on the interval and has the form	An exponential (exponential) is called the probability distribution of a continuous random variable X, which is described by a density having the form
		where λ is a constant positive value
distribution function
Probability hitting the interval
Expected value
Dispersion
Standard deviation

Examples of solving problems on the topic "Uniform and exponential laws of distribution"

Task 1.

Buses run strictly according to the schedule. Movement interval 7 min. Find: (a) the probability that a passenger coming to a stop will wait for the next bus for less than two minutes; b) the probability that a passenger approaching the stop will wait for the next bus for at least three minutes; c) the mathematical expectation and the standard deviation of the random variable X - the passenger's waiting time.

Solution. 1. By the condition of the problem, a continuous random variable X=(passenger waiting time) evenly distributed between the arrivals of two buses. The length of the distribution interval of the random variable X is equal to b-a=7, where a=0, b=7.

2. The waiting time will be less than two minutes if the random value X falls within the interval (5;7). The probability of falling into a given interval is found by the formula: P(x 1<Х<х 2)=(х 2 -х 1)/(b-a) .
P(5< Х < 7) = (7-5)/(7-0) = 2/7 ≈ 0,286.

3. The waiting time will be at least three minutes (that is, from three to seven minutes) if the random value X falls into the interval (0; 4). The probability of falling into a given interval is found by the formula: P(x 1<Х<х 2)=(х 2 -х 1)/(b-a) .
P(0< Х < 4) = (4-0)/(7-0) = 4/7 ≈ 0,571.

4. Mathematical expectation of a continuous, uniformly distributed random variable X - the passenger's waiting time, we find by the formula: M(X)=(a+b)/2. M (X) \u003d (0 + 7) / 2 \u003d 7/2 \u003d 3.5.

5. The standard deviation of a continuous, uniformly distributed random variable X - the passenger's waiting time, we find by the formula: σ(X)=√D=(b-a)/2√3. σ(X)=(7-0)/2√3=7/2√3≈2.02.

Task 2.

The exponential distribution is given for x ≥ 0 by the density f(x) = 5e – 5x. Required: a) write an expression for the distribution function; b) find the probability that, as a result of the test, X falls into the interval (1; 4); c) find the probability that as a result of the test X ≥ 2; d) calculate M(X), D(X), σ(X).

Solution. 1. Since, by condition, exponential distribution , then from the formula for the probability distribution density of the random variable X we obtain λ = 5. Then the distribution function will look like:

2. The probability that as a result of the test X falls into the interval (1; 4) will be found by the formula:
P(a< X < b) = e −λa − e −λb .
P(1< X < 4) = e −5*1 − e −5*4 = e −5 − e −20 .

3. The probability that as a result of the test X ≥ 2 will be found by the formula: P(a< X < b) = e −λa − e −λb при a=2, b=∞.
Р(Х≥2) = P(1< X < 4) = e −λ*2 − e −λ*∞ = e −2λ − e −∞ = e −2λ - 0 = e −10 (т.к. предел e −х при х стремящемся к ∞ равен нулю).

4. We find for the exponential distribution:

• mathematical expectation according to the formula M(X) =1/λ = 1/5 = 0.2;
• dispersion according to the formula D (X) \u003d 1 / λ 2 \u003d 1/25 \u003d 0.04;
• standard deviation according to the formula σ(X) = 1/λ = 1/5 = 1.2.

Recall the definition of the probability density.

We now introduce the concept of a uniform probability distribution:

Definition 2

A distribution is called uniform if, on an interval containing all possible values ​​of a random variable, the distribution density is constant, that is:

Picture 1.

Find the value of the constant $\ C$ using the following distribution density property: $\int\limits^(+\infty )_(-\infty )(\varphi \left(x\right)dx)=1$

\[\int\limits^(+\infty )_(-\infty )(\varphi \left(x\right)dx)=\int\limits^a_(-\infty )(0dx)+\int\limits ^b_a(Cdx)+\int\limits^(+\infty )_b(0dx)=0+Cb-Ca+0=C(b-a)\] \ \

Thus, the uniform distribution density function has the form:

Figure 2.

The graph has the following form (Fig. 1):

Figure 3. Density of uniform probability distribution

Uniform Probability Distribution Function

Let us now find the distribution function for a uniform distribution.

To do this, we will use the following formula: $F\left(x\right)=\int\limits^x_(-\infty )(\varphi (x)dx)$

1. For $x ≤ a$, according to the formula, we get:

1. For $a

1. For $x> 2$, according to the formula, we get:

Thus, the distribution function has the form:

Figure 4

The graph has the following form (Fig. 2):

Figure 5. Uniform probability distribution function.

Probability of a random variable falling into the interval $((\mathbf \alpha ),(\mathbf \beta ))$ under a uniform probability distribution

To find the probability of a random variable falling into the interval $(\alpha ,\beta)$ with a uniform probability distribution, we will use the following formula:

Expected value:

Standard deviation:

Examples of solving the problem for a uniform distribution of probabilities

Example 1

The interval between trolleybuses is 9 minutes.

Compile the distribution function and distribution density of the random variable $X$ waiting for the trolley bus passengers.

Find the probability that the passenger will wait for the trolleybus in less than three minutes.

Find the probability that the passenger will wait for the trolleybus in at least 4 minutes.

Find the mathematical expectation, variance and standard deviation

1. Since the continuous random variable $X$ of waiting for the trolleybus is uniformly distributed, then $a=0,\ b=9$.

Thus, the distribution density, according to the formula of the density function of the uniform probability distribution, has the form:

Figure 6

According to the formula of the uniform probability distribution function, in our case, the distribution function has the form:

Figure 7

1. This question can be reformulated as follows: find the probability that a random variable of a uniform distribution falls into the interval $\left(6,9\right).$

We get:

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