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Fundamental Theorem of Statistics

Statistics is an important branch of mathematics which deals with large numerical data that is applied to many different fields, such as commerce, science, industry etc. The study of gathering, presenting, interpreting, analyzing, managing and calculating data is known as statistics. There are various concepts that work as basics or fundamentals of statistics. In this article, we are going to learn about few important statistical topics, such as mean, standard derivation, central limit theorem and example problems based on them.

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Mean is perhaps the most fundamental concept in statistics. The mean is referred to the most commonly used measure of central tendency. The mean can also be equivalent to the average of given observations and therefore at times, mean is also termed as the arithmetic mean. When the discrete (not continuous) data is given, the mean is said to be the sum of all the values divided by the total number of values. In statistics, the terms mean, average, arithmetic mean, and mathematical expectation are said to be synonyms of the central value of a given discrete data set.

The mean of a data set is denoted by placing a bar over x, i.e. $\bar{x}$ (read as x bar), where x indicates mean of all given x's. Let us suppose that we are given a set of numbers denoted by $x_{1}, x_{2}, x_{3}..., x_{n}$ and the total number of values are eventually "n". Then, the mean of the given data set can be obtained by the following formula:

$\bar{x} = \frac{x_{1} + x_{2} + x_{3} + ... + x_{n}}{n}$

In the case, when the frequency (or weight) of each data value is given, then we find the mean using the formula given below:

$\bar{x} = \frac{x_{1} f_{1} + x_{2} f_{2} + x_{3} f_{3} + ... + x_{n} f_{n}}{n}$

This is known as weighted mean or weighted average.

Standard Deviation

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In statistics, the standard deviation is an important measure of dispersion. Standard deviation measures the amount of variation in the given set of data. It indicates the deviation of the data values from the central position. In other words, standard deviation represents how the values are spread or dispersed around the mean. It is abbreviated as SD and is denoted by $\sigma$ (population SD, read as sigma) and sometimes by s (sample SD). The following image illustrates the general spread of data when it is normally distributed.

Standard Deviation

The formulae for standard deviation are given below.

For Population
$\sigma$ = $\sqrt{\frac{\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}}{n}}$

For Sample
$\sigma$ = $\sqrt{\frac{\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}}{n-1}}$

Where $x_{i}$s are the given data values, n is total number of values and $\bar{x} is the mean of the data.

If the value of standard deviation is close enough to zero, then it means that given data values are spread closer to the mean. On the other hand, when the value of standard deviation is high, it indicates that the points spread quite far from the mean or they are dispersed over a wide range of values.

The Central Limit Theorem

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The central limit theorem states that as the size of the sample becomes larger, the sampling distribution of sampling means tend to be a normal distribution, irrespective of what the shape of population distribution is. Thus, if the sample size is above 30. the sample means would be normally distributed.

(1) The central limit has the following two statements. i.e.

     $\mu(\bar{x}) = \mu$

     where, $\mu$ denotes population mean and $\bar{x}$ denotes sample mean.

(2) The standard deviation of sample means is equal to the standard error of the population mean.

    $\sigma(\bar{x}) = \frac{\sigma}{\sqrt{n}}$

      where, $\frac{\sigma}{\sqrt{n}}$ is said to be the standard error.


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Few examples are discussed below.

Example 1: Calculate the mean of the following sample

6, 4, 8, 3, 7, 5, 3, 4

Solution: Formula for mean

$\bar{x} = \frac{x_{1} + x_{2} + x_{3} + ... + x_{n}}{n}$

Here, n = 8

$\bar{x} = \frac{6 + 4 + 8 + 3 + 7 + 5 + 3 + 4}{8}$

$\bar{x}$ = $\frac{40}{8}$ = $5$

Example 2: Calculate the standard deviation of following data: 3, 5, 4,10.

Solution: Calculation of mean

$\bar{x} = \frac{x_{1} + x_{2} + x_{3} + ... + x_{n}}{n}$

Given, $n$ = $4$

$\bar{x} = \frac{3 + 5 + 4 + 10}{4}$

$\bar{x}$ = $\frac{22}{4}$ = $5.5$

Table for standard deviation

$x_{i} - \bar{x}$
$(x_{i} - \bar{x})^{2}$
3 - 5.5 = - 2.5 
5 - 5.5 = -0.5  0.25 
4 - 5.5 = -1.5 2.25 
10  10 - 5.5 = 4.5  20.25 
    $\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}$ = 29

$\sigma$ = $\sqrt{\frac{\sum_{i=1}^{n}(x_{i} - \bar{x})^{2}}{n-1}}$

$\sigma$ = $\sqrt{\frac{29}{4-1}}$

$\sigma$ = $\sqrt{\frac{29}{4-1}}$

= $\sqrt{9.667}$ = $3.11$
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