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Data Analysis and Probability

Normal Distribution Function

Understanding Normal Distribution

Did you know that the height of students in your school, national exam scores, or even the weight of apples in the supermarket all follow the same pattern? This pattern is called normal distribution.

Normal distribution was first introduced by Abraham de Moivre in 17331733 as an approximation of the binomial distribution for large nn. Later, Pierre-Simon Laplace developed it further and it became known as the Moivre-Laplace Theorem. Laplace used normal distribution to analyze errors in experiments.

Normal distribution is very useful because many natural and social phenomena follow this pattern. From human height to scientific measurement results, everything tends to be normally distributed.

Characteristics of Normal Curve

Imagine a very symmetrical hill, like an inverted bell. That's the shape of the normal distribution curve. Let's look at the visualization of a standard normal curve with μ=0\mu = 0 and σ=1\sigma = 1:

Standard Normal Distribution Curve
Bell-shaped curve that is symmetrical with characteristic features of normal distribution.

This curve has several unique characteristics that make it special:

  1. Shape and Symmetry

    The curve is bell-shaped and symmetrical about the vertical line passing through the mean (μ\mu). This means the left and right parts of the curve are perfect mirrors of each other.

  2. Central Point

    The mean, median, and mode are all located at the same point, which is at the peak of the curve. This occurs because the distribution is symmetrical.

  3. Inflection Points

    The curve has inflection points at x=μ±σx = \mu \pm \sigma, which means the curve changes from concave to convex (or vice versa) at a distance of one standard deviation from the mean.

  4. Horizontal Asymptote

    The curve approaches the xx axis but never touches it, both at the left and right ends.

Its Mathematical Function

Now, let's look at its mathematical formula. Don't worry if it looks complicated, the important thing is that you understand the concept.

If XX is a normal random variable with mean μ\mu and variance σ2\sigma^2, then the normal distribution function can be written as:

f(x;μ,σ)=1σ2πe12(xμσ)2f(x; \mu, \sigma) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}

for <x<-\infty < x < \infty.

Where:

  • π\pi is the constant 3.14163.1416
  • ee is the constant number 2.71832.7183
  • μ\mu is the mean of the distribution
  • σ\sigma is the standard deviation

This formula looks complex, but you don't need to memorize it or calculate it manually. The important thing is to understand that the shape of the curve is determined by the values of μ\mu and σ\sigma.

Transformation to Standard Normal

In practice, we often use the standard normal distribution with mean μ=0\mu = 0 and standard deviation σ=1\sigma = 1. To convert a regular normal distribution to standard normal, we use the transformation:

Z=xμσZ = \frac{x - \mu}{\sigma}

This variable ZZ is called the standard score or z-score. This transformation is very useful because it allows us to use the readily available standard normal distribution table.

Why use z-score?

With this transformation, we can compare data from different distributions. For example, you can compare math scores with physics scores, even though their means and standard deviations are different.

Calculation Example

Let's look at a practical example. Suppose we have a normal distribution with μ=50\mu = 50 and σ=10\sigma = 10. We want to find the probability that XX lies between 4545 and 6262.

Step 1: Transform to z-score

z1=455010=510=0.5z_1 = \frac{45 - 50}{10} = \frac{-5}{10} = -0.5
z2=625010=1210=1.2z_2 = \frac{62 - 50}{10} = \frac{12}{10} = 1.2

Step 2: Use the standard normal distribution table

We need to find P(0.5<Z<1.2)P(-0.5 < Z < 1.2). Remember that for interval probability, we use the formula:

P(a<Z<b)=P(Z<b)P(Z<a)P(a < Z < b) = P(Z < b) - P(Z < a)

From the standard normal distribution table:

  • P(Z<0.5)=0.3085P(Z < -0.5) = 0.3085
  • P(Z<1.2)=0.8849P(Z < 1.2) = 0.8849

Step 3: Calculate the final probability

P(45<X<62)=P(0.5<Z<1.2)=0.88490.3085=0.5764P(45 < X < 62) = P(-0.5 < Z < 1.2) = 0.8849 - 0.3085 = 0.5764

So, the probability that XX lies between 4545 and 6262 is 0.57640.5764 or approximately 57.64%57.64\%.