Proof of Normal Distribution Formulas

We assume that X is a normal random variable or X is normally distributed, with parameters π and σ² if the probability density function for X is

This density function is a clapper-shaped curve that is approximately µ (see Figure 1). The values of μ and σ² respectively represent, in a certain sense, the average value and diversity that might occur at X.

The normal distribution was first introduced by the French mathematician Abraham De Moivre in 1733 and was used by him to approach opportunities related to the binom probability distribution if the binom parameter n is large. This finding was later extended by Laplace and others and is now included in the opportunity theory called the central limit theorem, which will be discussed in the next article. This central limit theorem, one of two important findings in probability theory (the other is a strong law of large numbers), provides a theoretical basis for empirical observations that are often discovered, namely that many random phenomena follow, at least approach, a normal distribution of opportunities . Some examples are height, speed, in any direction from a gas molecule in space, and errors that occur when measuring a physical quantity.

To prove that f(x) is indeed a density function of opportunity, we need to understand the derivative and integrative functions that show that

By substituting y = (x-μ) / σ, we get dx = σ dy,

Then,

So, we have to show that

For this purpose, for example

Then

Now, we will calculate the double integral above through the exchange of variables to the polar coordinates. For example x = r cos Ɵ, y = r sin Ɵ , dy dx = r dƟ dr. Then,

Because cos²Ɵ + sin²Ɵ = 1 then y² + x²

y² + x² = r²

first, we look for an integral of Ɵ,

Then

For Example u = r²/2 and du = r dr Then

So, I = √2π and prove that

An important fact about the normal probability distribution is that if X is normally distributed with parameters μ and σ2, then Y = αX + β is normally distributed with parameters αμ + β and α²σ². This is due to the distribution function F_y, the cumulative distribution function for the probability distribution Y, given by,

In this case that equation (1) is obtained through the variable y = αx + β. However, because

, then from equation 1 it is obtained that the opportunity density function for Y, namely f_r(y), is

So, Y is spread normally with parameters αμ + β and (ασ)².

An important implication of the above results is that if X spreads normally with parameters μ and σ², then Z = (X – μ)/σ also spreads normally but with parameters 0 and 1. The probability distribution of Z thus is said to have a standard normal distribution.

Already, it is customary to symbolize the cumulative distribution function for the standard normal probability distribution as Φ(x). So,

the values of Φ(x) for non-negative x are listed in Table 1 below. For negative x values, the value Φ(x) can be obtained from equation (2).

Φ(- x) = 1 - Φ(x) - ∞ < x < ∞ … … … … … (2)

Proof of Equation (2) which is a result of the embrace of standard normal density functions, is provided as an exercise. This equation says that if Z is a standard normal probability distribution, then

P{Z ≤ -x} = P{Z > x} - ∞ < x < ∞

Because, Z = (X – μ)/σ is a standard normal probability distribution if X is spread normally with parameters μ and σ², the distribution function for X can be written as,

Example 1.

If X is a normal probability distribution with parameters μ = 3 and σ² = 9, calculate it

P{1 < X < 5}, P{X > 0}, dan P{|X-3| > 6}.

Answer:

Note: Φ(- x) = 1 - Φ(x)

the procedure to get the value of Φ(1/3)= Φ(0.33), Φ(2/3)= Φ(0.66), Φ(1), note in the following Table 1,

Example 2.

An exam is often considered good (in the sense of giving a scattered score for those who take it) if the resulting test score can be approached by a normal density function. (In other words, the frequency score graph has a clapper shape). Lecturers often use test score scores to infer normal parameters μ and σ², and then assign a value of A to those who score greater than μ + σ, B for those whose values are between μ and μ + σ, C for those whose values are between μ - σ dam μ, D for those whose value is between μ - 2σ and μ - σ, and F for those whose value is below μ - 2σ. Because,

Then about 16 percent of the examinees will get an A, 34 percent B, 34 percent C, 14 percent D, and 2 percent F.

Example 3.

An expert witness in the trial said that the duration (in days) of pregnancy (from conception to the birth of a baby) spread approximately normally with parameters μ = 270 and σ² = 100. The defendant successfully presented evidence that he was abroad during the period that began 290 days before the baby's birth to 240 days before the baby's birth. If he really is the baby's father, what is the probability that the mother has a very long or very short pregnancy as indicated by the defendant's confession?

Solution: Let X state the length of the pregnancy and assume that the defendant is indeed the father of the baby in the womb. Then the probability that birth can occur in that period is?

P{Z ≤ -x} = P{Z > x}

P{-x ≥ Z} = P{x > Z}