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The average calculator helps find a data set's average or arithmetic mean. It also shows the calculation steps and other important statistics.
Average
Sum
Count
=
389
8
=
48.625
Sum | 389 | Largest | 234 |
---|---|---|---|
Count | 8 | Smallest | 2 |
Median | 23 | Range | 232 |
Geometric Mean | 22.87894539 |
There was an error with your calculation.
The online average calculator makes it easy to find the average for any data set. You can type, copy, and paste your data into the data box. Make sure to separate each data point with a comma. Then, click the "Calculate" button.
The average calculator will show you the average (arithmetic mean), calculation steps, & other related statistics for the data set.
The average is defined as the mean of the values in a data set. All the values in the data set are used to calculate the average. Therefore, it represents the entire data set. The average is regarded as one of the most important central tendency or summary measures.
The simple arithmetic mean is the most common average. However, there are several kinds of averages, including the geometric mean, weighted average, combined arithmetic average, harmonic mean, and so forth.
The average of a population is represented by μ (Mu) and the average of a sample is represented by X̄ (X bar).
The simple average is calculated by dividing the data set's values by the total number of data items. The simple average is sometimes referred to as the mean, the arithmetic mean, and the average.
To calculate the average of a population, we can use the formula below.
μ = Sum of the data set’s values / Total number of data values in the population = ΣX / N
To calculate the average of a sample, we can use the below formula:
X̄ = Sum of the data set’s values / Total number of data values in the sample = ΣX/n
Let's learn the average using the below example.
Example
Jasmine's scores for seven subjects from the previous semester are displayed in the table below. What is the average of Jasmine's previous semester subject scores?
Subject | Score |
---|---|
Management | 84 |
Communication | 90 |
Accounting | 75 |
Economics | 60 |
Business Statistics | 85 |
International studies | 92 |
Mathematics | 81 |
Solution
The average score = ΣX / N = (84 + 90 + 75 + 60 + 85 + 92 + 81) / 7 = 567 / 7 = 81
The average is a concept everyone is familiar with. The average income, the average cost of production, average pricing, average score, average fuel consumption, etc., are a few examples you may have heard often. Even in everyday life, the simple average is a standard computation. The simple average or the simple arithmetic mean is also known as the ideal average.
In some situations, however, we use other measures of central tendency. Let's take a look at them.
The arithmetic mean is not an appropriate measurement when determining the average growth rate of a value over time. The geometric mean, which is often used in accounting and finance, such as in calculating compound interest, is a much better indicator for such calculations. This is because the growth rate is multiplicative rather than additive.
The geometric mean of your data set is defined as the nth root of the product of n items. It is calculated by multiplying each value together and then calculating the nth root of the product, where n is the number of items in the dataset. The geometric mean is helpful when averaging ratios, percentages, and growth rates.
$$Geometric\ Mean = \sqrt[n]{x₁×x₂×x₃×…×xₙ} = (x₁×x₂×x₃×…×xₙ)^{\frac{1}{n}}$$
We will find the Geometric Mean of the previous example.
$$Geometric\ Mean = \sqrt[7]{84×90×75×60×85×92×81} = 80.31$$
The Geometric Mean is always equal to or lower than the simple average (arithmetic mean).
In our example,
Geometric Mean ≤ The average
80.31 < 81
You can use the average calculator to determine more than just the arithmetic mean. You can also use it to obtain your data set's Geometric Mean.
In the simple arithmetic mean, all values have the same weight or importance. But in some cases we cannot apply the same level of importance to every value in our dataset.
In our example, we calculated the average by summing up all the scores and dividing by the total number of subjects. We haven't considered the relative importance of each subject.
The weighted average must be used when we need to consider the relative importance of each item of our data set when calculating the average. The weighted average is calculated by dividing the weighted values by the total of the weights. The data value multiplied by the relevant weight is the weighted value.
We can use the below formula to find the weighted average.
The weighted average = The sum of the weighted values / The sum of the weights = ΣWX / ΣW
Example
Assume that each of the subjects in the previous example has a different weight. So, the updated data table for Jasmine's score in 7 subjects of the prior semester is as follows.
Weighted average of Jasmine's scores from the previous semester
Subject | Score | Weight |
---|---|---|
Management | 84 | 3 |
Communication | 90 | 2 |
Accounting | 75 | 4 |
Economics | 60 | 3 |
Business Statistics | 85 | 3 |
International studies | 92 | 2 |
Mathematics | 81 | 3 |
Solution
The weighted average score = ΣWX / ΣW = (84×3+90×2+75×4+60×3+85×3+92×2+81×3)/(3+2+4+3+3+2+3) = (252+180+300+180+255+184+243)/20 = 1594/20 = 79.7
The median is the midway value of a data collection when it is arranged ascending (lowest value to highest value) or descending (highest value to the lowest value). In other words, the median is the point at which the data array (An array is an arrangement of raw data in ascending or descending order of values) is divided into 2 equal parts. As a result, 50% of the values are below the median, and 50% are above the median.
When finding the median first, we have to find the median's position using the formula below:
$$The\ position\ of\ the\ median = \left( \frac{n+1}{2} \right)^{th}item$$
The "n” denotes the overall item count of the data set.
If the total number of items in the dataset is odd, the value of the item at the center position is the median. But suppose the total number of items in the data set is an even figure. In that case, the average between the two numbers in the middle is the median.
The mean, or average, is calculated by summing all the values in a data set and then dividing by the number of observations. It gives us a value that considers each point in the data set. In contrast, the median is the middle value in a data set ordered from lowest to highest and provides a central point that divides the data set in half, but does not take into account the magnitude of all values.
Both the mean and median can be visually estimated from a graphical representation of data. The mean can be roughly estimated in a symmetric distribution as it should lie at the center, while the median can be determined as the middle value in a box plot, for example.
Both the mean and median have their uses in further statistical analysis. The mean is particularly useful for data that is normally distributed and does not contain outliers, as it is included in calculations of variance and standard deviation. The median is valuable as a measure of central tendency when the data is skewed or contains outliers, and it is frequently used in non-parametric statistical tests that do not assume a specific data distribution.
The mean is the most suitable measure of central tendency when the data set has a symmetric distribution without outliers. It is a reliable indicator of the center of the data because it incorporates every value. If a data set does contain outliers, it may be preferable to remove these before calculating the mean to ensure an accurate representation of the central tendency.
The median is the preferred measure of central tendency when dealing with skewed distributions or when outliers are present. This is because the median, being the middle value of a data set ordered from lowest to highest, is not influenced by extreme values, unlike the mean. In such cases, the median provides a better central value that represents the majority of the data without being distorted by outliers.
Let's modify our original example and learn about the outliers.
Example
Assume Jasmine received 15 for international studies instead of 92. What is the average of Jasmine's new scores from the previous semester's subjects?
Subject | Score |
---|---|
Management | 84 |
Communication | 90 |
Accounting | 75 |
Economics | 60 |
Business Statistics | 85 |
International studies | 15 |
Mathematics | 81 |
Solution
The average score = ΣX / N = (84+90+75+60+85+15+81)/7 = 490/7 = 70
The new average score is 70. It is reduced from 81 to 70 by 11. You were able to see how the outliers affect the average.
In this kind of situation, the median of the data is a more suitable central tendency measure than the mean. To understand this, let's calculate the median for the original and modified examples.
Example
The table below displays Jasmine's original score for seven subjects from the previous semester. What is the median of Jasmine's previous semester subject scores?
Subject | Score |
---|---|
Management | 84 |
Communication | 90 |
Accounting | 75 |
Economics | 60 |
Business Statistics | 85 |
International studies | 92 |
Mathematics | 81 |
Solution
As the first step, we will arrange all scores as an array. Depending on your preference, you can organize it in ascending or descending order.
60, 75, 81, 84, 85, 90, 92
$$The\ position\ of\ the\ median = \left( \frac{n+1}{2} \right)^{th}item = \left( \frac{7+1}{2} \right)^{th}item = 4^{th}item$$
Next, we will check what is the 4th item of our data set. It is 84. Therefore, the median of the data set is 84. Now, we will find the median of the modified data set with the outliers.
Example
Assume Jasmine received 15 instead of 92 for international studies. What is the new median score for the subjects Jasmine took last semester?
Subject | Score |
---|---|
Management | 84 |
Communication | 90 |
Accounting | 75 |
Economics | 60 |
Business Statistics | 85 |
International studies | 15 |
Mathematics | 81 |
Solution
As the first step, we will arrange all scores as an array. Let’s arrange our data in ascending order.
60, 75, 81, 84, 85, 90, 92
$$The\ position\ of\ the\ median = \left( \frac{n+1}{2} \right)^{th}item = \left( \frac{7+1}{2} \right)^{th}item = 4^{th}item$$
Now, we will check what is the 4th item of our data set. It is 84 and represents the data set's median.
Even though there is an outlier in this case, the median has not been affected.