Question

Find an example of application of Normal Distribution (or approximately Normal Distribution) in your workplace or business or any example of Normal Distribution. Prove that the variable has the characteristics of a Normal Distribution. Recall that the variable must be continuous and the distribution must be symmetrical (or approximately symmetrical). To prove that the distribution is approximately symmetrical select 20 random observations (measurements/data) of the variable and run a Descriptive Statistics using your calculator or Excel. Just copy the output (results) to your posting. In the Descriptive Statistics, if the Mean and Median are fairly close and the skewness is close to zero (under 1), you can conclude that the distribution is approximately symmetrical.

Answer 1

normal distributin:

The normal distribution is commonly associated with the 68-95-99.7 rule which you can see in the image above. 68% of the data is within 1 standard deviation (σ) of the mean (μ), 95% of the data is within 2 standard deviations (σ) of the mean (μ), and 99.7% of the data is within 3 standard deviations (σ) of the mean (μ).

This post explains how those numbers were derived in the hope that they can be more interpretable for your future endeavors. As always, the code used to derive to make everything (including the graphs) is available on my github. With that, let’s get started!

Probability Density Function

To be able to understand where the percentages come from, it is important to know about the probability density function (PDF). A PDF is used to specify the probability of the random variable falling within a particular range of values, as opposed to taking on any one value. This probability is given by the integral of this variable’s PDF over that range — that is, it is given by the area under the density function but above the horizontal axis and between the lowest and greatest values of the range. This definition might not make much sense so let’s clear it up by graphing the probability density function for a normal distribution. The equation below is the probability density function for a normal distribution

PDF for a Normal Distribution

Let’s simplify it by assuming we have a mean (μ) of 0 and a standard deviation (σ) of 1.

PDF for a Normal Distribution

Now that the function is simpler, let’s graph this function with a range from -3 to 3.

hight
117.9
106.3
103.3
105.9
116.7
107.4
105.8
106.3
99.4
108.7
104.1
110.2
111
106.9
110
103.5
114.9
104.5
108.1
105.7

Discriptive Statistics :

Column1
Mean	107.83
Standard Error	1.032016
Median	106.6
Mode	106.3
Standard Deviation	4.615318
Sample Variance	21.30116
Kurtosis	0.493446
Skewness	0.733533
Range	18.5
Minimum	99.4
Maximum	117.9
Sum	2156.6
Count	20
Largest(1)	117.9
Smallest(1)	99.4
Confidence Level(95.0%)	2.160035

above observation of mean , median ,mode is approximately equal so this data set follows approximately Normal Distribution.