Normal Distribution Density Function

Normal Distribution

Measurements such as human height or repeated laboratory results often cluster around a central value. In some populations and under suitable conditions, a normal distribution can approximate that pattern. Check the model against the observed data before using it.

In $1733$ , Abraham de Moivre derived a normal approximation to the binomial distribution for large $n$ . Pierre-Simon Laplace later extended this result, which is now associated with the de Moivre-Laplace theorem and the development of the central limit theorem.

The normal distribution is one possible model. Use it when observations are roughly symmetric around a mean and the model fits the data well enough for the question being studied.

Characteristics of Normal Curve

The normal density has a symmetric bell shape. The visualization below shows the standard normal curve with $\mu = 0$ and $\sigma = 1$ :

Standard Normal Distribution Curve

The standard normal density is centered at

\mu=0

. The vertical segments mark its inflection points

\mu\pm\sigma

The curve has four defining characteristics:

Shape and Symmetry

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

The normal density is symmetric and has one peak. Therefore, its mean, median, and mode all coincide at $\mu$ .
Inflection Points

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

Probability Density Function

The formula below gives the probability density at each value of $x$ .

If $X$ is normally distributed with mean $\mu$ , variance $\sigma^2$ , and $\sigma>0$ , its density function is:

f_{\mu,\sigma}(x) = \frac{1}{\sigma\sqrt{2\pi}} e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}

This formula applies to every $x$ satisfying $-\infty < x < \infty$ .

The constants and parameters are:

$\pi \approx 3.1416$ is the circle constant
$e \approx 2.7183$ is Euler's number
$\mu$ is the mean of the distribution
$\sigma>0$ is the standard deviation

The parameter $\mu$ shifts the center of the curve, while $\sigma$ controls its spread. Find probabilities from areas under the density curve. The height $f(x)$ at one point is a density value.

Transformation to Standard Normal

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

Z = \frac{x - \mu}{\sigma}

The variable $Z$ is the standard score, or z-score. It measures how many standard deviations $x$ lies above or below the mean. Standardization maps every normal distribution to the same $Z$ scale, so one standard normal table works for all of them.

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.

Finding a Normal Probability Between Two Bounds

Suppose $X$ is normally distributed with $\mu = 50$ and $\sigma = 10$ . Find the probability that $X$ lies between $45$ and .

Step 1: Transform to z-score

z_1 = \frac{45 - 50}{10} = \frac{-5}{10} = -0.5

Step 2: Use the standard normal distribution table

The transformed interval is $P(-0.5 < Z < 1.2)$ . Subtract the cumulative probability at the lower boundary from the cumulative probability at the upper boundary:

\begin{aligned} P(a < Z < b) &= P(Z < b) \\ &\quad{}- P(Z < a) \end{aligned}

From the standard normal distribution table:

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

Step 3: Calculate the final probability

\begin{aligned} &P(45 < X < 62) \\ &= P(-0.5 < Z < 1.2) \\ &= 0.8849 - 0.3085 \\ &= 0.5764 \end{aligned}

So, the probability that $X$ lies between $45$ and $62$ is $0.5764$ or approximately $57.64\%$ .