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:
| 157 | 152 | 151 | 155 | 163 | 134 | 141 | 149 | 165 | 144 |
| 147 | 148 | 145 | 150 | 139 | 143 | 146 | 156 | 148 | 164 |
| 166 | 152 | 142 | 141 | 151 | 131 | 153 | 134 | 161 | 156 |
| 130 | 145 | 163 | 140 | 173 | 150 | 164 | 152 | 137 | 141 |
| 154 | 159 | 170 | 139 | 165 | 131 | 138 | 157 | 133 | 166 |
| 154 | 140 | 135 | 146 | 142 | 156 | 145 | 160 | 141 | 136 |
| 157 | 171 | 145 | 148 | 141 | 161 | 153 | 165 | 152 | 147 |
| 166 | 156 | 159 | 154 | 138 | 150 | 156 | 150 | 166 | 149 |
| 163 | 141 | 138 | 167 | 143 | 152 | 144 | 155 | 157 | 149 |
| 157 | 144 | 156 | 152 | 175 | 141 | 154 | 133 | 155 | 146 |
| Table 1. Initial data for checking the normality of the distribution | |||||||||
| # | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
|---|---|---|---|---|---|---|---|---|---|---|
| x | 7 | 6 | 15 | 12 | 18 | 15 | 10 | 8 | 6 | 2 |
| pi | 0.07 | 0.06 | 0.15 | 0.12 | 0.18 | 0.15 | 0.1 | 0.08 | 0.06 | 0.02 |
| Table 2. Number of elements in each interval | ||||||||||
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 = 150.52
σ = 10.37
The calculation of the mean and standard deviation is described in the article distribution parameters
Normal distribution
The normal distribution curve for μ=150.52 and σ=
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:
P(x) = e^[-0.5((x-150.52)/10.37)2] / [10.37√2π] Normal distribution formula
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 = 17.94
S = 91.76
Δ = 20%
The deviation is 20%, so i conclude that the distribution is normal
according to the normality criterion with an average value
μ=150.52 and standard deviation σ=
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.
