So I understand the concept of a distribution. I just don’t understand the concept of a standard deviation from the mean of that distribution. How can we tell what is 1 standard deviation away as opposed to 2 standard deviations?
I’d be very thankful for an explanation, thank you :).
In: 9
I’ll explain this the way I understood it, so ill break it down to its simplest and most manual method.
Since you understand distribution, picture a variable distributed on a single plot line, a simple straight line. Let’s say you were trying to find the standard deviation of the weight of students in a class, for simplicity. The weights will be distributed on the line ranging from 50kg til 120kg, for instance, with a mean of 75kg. With the mean, you know the average, or “center” weight in terms of the plot line.
The problem with the average or mean, is that it does not describe how far a single weight can actually be from the mean. This is relevant if you want to make sure your population (the variable of your study) is not too far off from the value of the mean (like in production of materials where precision of thickness, height, weight, etc is important, you would want your entire batch to be near the mean value).
For example, you can find a mean of 75 if there were three students, one being 50kg, one being 75kg, and one being 100kg. But you could also find a mean of 75kg if the three students were 50kg, 50kg, and 125kg.
Now, you can find the difference between the weight of each student and the mean weight. On the plot line, this difference would be represented by the distance of each weight from the mean. Note the distance between all values and the mean.
Now you can find the AVERAGE DISTANCE between the values and the mean. A high average distance indicates one of two things. Either your values are generally spread out and far from the mean, OR, that there is an outlier value affecting your average. An outlier is basically a single value that is so far off the mean that its affect on the average is considerable. Take for example the 125kg student in the second example. Due to this high value, the mean is 75kg even though majority of the students weigh less in comparison.
Additionally, note that if you were to find the average of those distances using conventional means, your result would be zero. For instance, using the example, the distances are 25 (since 75-50), 25 (75-50), and -50 (75-125). Then 25 + 25 – 50 = 0
To bypass this, we square all values individually (square of a negative value is positive) before finding the average.
That is standard deviation to my understanding.
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