How does central tendency relate to normal distribution?

1 Answer
Oct 5, 2017

Any normal distribution has a graph that is perfectly symmetric about a vertical line through its peak. Therefore, all measures of central tendency (most commonly, the mean, median, and mode) give the same answer: the #x#-value of the peak.

Explanation:

A normal distribution with mean #mu# and standard deviation #sigma# has formula #f(x)=1/(sigma sqrt{2pi})e^(-(x-mu)^2/2)#.

The graph of this function is the classical "bell-shaped curve". The standard normal distribution with #mu=0# and #sigma=1# is shown below.

graph{(1/sqrt(2pi))*e^(-x^2/2) [-2.5, 2.5, -1.25, 1.25]}

The mean of a general normal distribution is equal to the improper integral

#int_{-infty}^{infty}xf(x)dx=1/(sigma sqrt{2pi}) int_{-infty}^{infty}xe^(-(x-mu)^2/2)dx#.

This integral is not too hard to do when #mu=0# (use a substitution #u=-x^2/2# in that case). It can't be done with elementary functions when #mu!=0#, but other techniques (moment generating functions ) can still be used to show it equals #mu# in that case (in other words, the mean is the mean!! lol!!...see above for the name of #mu#).

The median is the value of #M# such that #int_{-infty}^{M}f(x)dx=1/2#. The convergence of the integral and the symmetry of the graph about the vertical line at #x=mu# can be used to say #M=mu# as well.

The mode is the #x#-value of the global maximum of this graph. Setting #f'(x)=0# and solving for #x# implies this global maximum occurs at #x=mu# (you should check this). The mode is therefore #x=mu# as well.

All the measures of central tendency therefore give the same answer: the #x#-value of the peak of the graph.