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Normality of the distribution

In the examples in this article, data is generated every time the page loads. If you want to see an example with different values - reload the page.

Some statistical tools assume that the distribution is normal. The algorithm for checking the normality of the distribution will be given below, and also an example in excel.

Distribution law

Checking for compliance with the normal distribution is a special case of solving the problem on finding among the known distribution functions one that describes as accurately as possible this distribution.

First of all, it is necessary to structure the available values, in the article properties distributions it describes how the distribution series is constructed, so here I will omit the details and give source data and processed values:

148 137 152 149 160 148 158 152 140 136
155 140 142 155 156 136 152 140 151 143
136 150 158 165 141 155 166 159 156 154
155 149 163 145 134 161 154 143 149 160
144 156 137 150 137 154 152 161 147 162
159 161 147 129 153 147 163 152 147 159
127 156 149 146 138 141 163 152 161 149
155 160 137 153 144 162 160 143 142 166
145 165 159 161 165 145 159 164 164 144
151 152 150 164 154 149 128 167 129 164
Table 1. Initial data for checking the normality of the distribution
# 12345678910
x4187101515111612
pi0.040.010.080.070.10.150.150.110.160.12
Table 2. Number of elements in each interval
Graph 1. Distribution range

Regardless of what we see on the graph, we need to check whether whether the distribution is normal.

The characteristics of a normal distribution are the mean and standard deviation. Let's calculate these values for our distribution:

μ = 151.03
σ = 9.7
The calculation of the mean and standard deviation is described in the article distribution parameters

Normal distribution

The normal distribution curve for μ=151.03 and σ=
Warning: Undefined variable $variation in /var/www/content/ktree/t9n/en/articles/statistics_check_is_normal.php on line 148
:

P(x) = e^[-0.5((x-151.03)/9.7)2] / [9.7√2π] Normal distribution formula
Graph 2. Distribution series and normal distribution, μ = 151.03, σ = 9.7

First approximation

Let's try to invent a criterion of normality, the simplest, what comes to mind is to determine the percentage of compliance the normal curve and the existing distribution.

To do this, add up the absolute values of the differences across all points of the graph, find the area under the normal distribution graph and calculate the deviation of interest, I will call such a criterion "criterion of normality" and I will decide that if the deviation more, let's say 30%, then the distribution is not normal.

diff = Σ|D(X) - P(X)|
S = ΣP(X)
Δ = diff / S
diff = 28.78
S = 90.83
Δ = 32%

The deviation is 32%, therefore, I draw the following conclusion: the distribution is not normal according to the normality criterion.

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