The probability density function of the normal distribution and its properties are presented starting from the probability histograms .
A normally distributed data set with mean \( \mu = 3.5 \) and a standard deviation \( \sigma = 1 \) is used to highlight
the link between the probability histogram of the data and the normal density function which leads to the definition of normal distribution .
Graphs of normal distributions are presented to highlight the effects of the mean \( \mu \) and the standard deviation \( \sigma \) on the normal distribution.
The properties of the normal distributions are presented. The standard normal distribution is defined and the steps used to go from normal distributions to standard normal distribution using the z-score are presented.
Examples of calculations of probabilities of normal distributions are included.
A PDF table as well a Google sheet for standard normal distribution area values are included and both may be downloaded and used in calculations.
In figure 1, are shown the probability histograms of 3 data sets. The data in histogram A is more concentrated to the left. The data in histogram B is more concentrated to the right; and the data in histogram C is spread out.
Figure 1
\( \mu \approx 3.5 \) , median \( \approx 3.5 \)
Figure 2
The distribution is symmetric and will be discussed in more detail below.
Figure 3
Figure 4
Figure 5
The probability density function of a normal distribution with mean \( \mu \) and standard deviation \( \sigma \) is defined by \[ \boxed {\displaystyle f_X(x) = \dfrac{1}{ \sigma \sqrt{2 \; \pi }} \; e^{-\frac{1}{2} \left( \dfrac{x - \mu}{\sigma} \right)^2 } } \] where \( X \) is the normally distributed random variable.
In figure 6 below, are shown normal probability density functions with the same standard deviation \( \sigma = 2 \) and different means.
Figure 6
Figure 7
1-The normal distribution is centered around the mean; this is clearly shown in figures 6 and 7.
2 - The mean and median of a normal distribution are very close (equal in theory).
3 - The total area under the curve of a normal distribution is equal to \( 1 \).
\[ \displaystyle \dfrac{1}{ \sigma \sqrt{2 \; \pi }} \int_{-\infty}^{\infty} \; e^{-\frac{1}{2} \left( \dfrac{x - \mu}{\sigma} \right)^2 } dx = 1 \]
4 - Data Distribution is as follows
a) Approximately, \( 68\% \) of the data lies within 1 standard deviation from the mean.
\[ \displaystyle \frac{1}{\sigma\sqrt{2\pi}} \int _{\mu-\sigma}^{\mu+\sigma}\:e^{-0.5\left(\frac{x-mu}{\sigma}\right)^2}dx \approx 0.68268\]
Figure 8
b) Approximately, \( 95\% \) of the data lies within 2 standard deviations from the mean.. \[ \frac{1}{\sigma\sqrt{2\pi \:}}\: \int _{\mu-2\sigma}^{\mu+2\sigma}\:e^{-0.5\left(\frac{x-mu}{\sigma}\right)^2}dx \approx 0.95449 \]
Figure 9
c) Approximately, \( 99\% \) of the data lies within 3 standard deviations from the mean.. \[ \frac{1}{\sigma\sqrt{2\pi \:}}\: \int _{\mu-3\sigma}^{\mu+3\sigma}\:e^{-0.5\left(\frac{x-mu}{\sigma}\right)^2}dx \approx 0.99730 \]
Figure 10
The normal distribution with mean \( \mu = 0 \) and standard deviation \( \sigma = 1 \) is called standard normal distribution and its probability density function is given by
\[ f_X(x) = \dfrac{1}{ \sqrt{2 \; \pi }} \; e^{-\frac{1}{2} x^2 } \]
The cumulative probability distribution of the standard normal distribution is defined by
\[ \displaystyle F_{X} (x) = \dfrac{1}{ \sqrt{2 \; \pi }} \int_{-\infty}^{x} \; e^{-\frac{1}{2} t^2} \; dt \]
and is used to find probabilities of the form
\[ P( X \le a) = F_{X} (a) \]
Figure 11
Hence \( P( X \le a) \) is given by the area between x-axis, the curve of the standard normal distribution and \( x = a \)
Figure 12
The normal distribution of mean \( \mu \) and standard deviation \( \sigma \) is given by
\[ \displaystyle f_{X}(x) = \frac{1}{\sigma\sqrt{2\pi \:}}\:e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2} \]
The probability \( P( X \le a) \) is given by the area between the x-axis, the curve of the normal distribution and \( x = a \) and is given by
\[ \displaystyle P( X \le a) = F_{X} (a) = \frac{1}{\sigma \sqrt{2\pi \:}}\:\int_{-\infty}^{a} \; e^{- \frac{1}{2} \left(\frac{t - \mu}{\sigma}\right)^2} \; dt \]
Let us use the substitution in the integral
\( z = \frac{t - \mu}{\sigma} \) which gives \( \dfrac{dz}{dt} = \dfrac{1}{\sigma} \) and substitute in the above integral
\[ \displaystyle P( X \le a) = F_{X} (a) = \frac{1}{\sqrt{2\pi \:}}\:\int_{-\infty}^{\frac{a - \mu}{\sigma}} \; e^{- \frac{1}{2} z^2} \; dz \]
We are now dealing with the integral of the standard normal distribution and z-score given by.
\[ z = \frac{a - \mu}{\sigma} \]
The above result tells us that you only need to know the integral of the standard normal distribution in order to calculate any probability related to any normal distribution and that is by using the z-score defined above.
The integral
\[ \frac{1}{\sqrt{2\pi \:}}\:\int_{-\infty}^{z_0} \; e^{- \frac{1}{2} z^2} \; dz \]
can be calculated using Google sheets that may be downloaded for personal use.
and is also given in the form of
Table of Normal Distribution in pdf may be downloaded and used.
Example 1
A random variable \( X \) is normally distibuted with a mean \( \mu = 2.2 \) and a standard deviation \( \sigma = 2.5 \). Find the probability
a) \( \qquad P( X \le 1.2) \)
Solution to Example 1
In this example we have \( \mu = 2.2 \) and \( \sigma = 2.5 \), hence the z-score defined above is given by
\[ z = \frac{1.2 - 2.2}{2.5} \approx -0.4 \]
Different ways to calculate the integral
a) Using a calculator, the probability is given by
\[ P( X \le 1.2) = \frac{1}{\sqrt{2\pi \:}}\:\int_{-\infty}^{-0.4} \; e^{- \frac{1}{2} z^2} \; dz \approx 0.34457 \]
b) Use a table of values of the probability of a standard normal distribution which may be
downloaded for personal use.
Figure 13
NOTE a Table of Normal Distribution in pdf may be downloaded and used.
