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December 25, 201539 Insightful Publications

Arithmetic mean is one of the most important chapters of Maths. It is introduced in lower grades and is referred to as average however, in 10th boards, students are taught different approaches to calculate the arithmetic mean. Statistics is a vital part of the syllabus in 12th boards and students need to have basic knowledge of arithmetic mean to be able to attend the sums appropriately. This article will include all the details like definition, properties, formulae and examples related to the chapter of arithmetic mean. Follow this page to get a clear idea of the concepts related to the chapter of arithmetic mean.

Students need to practice a significant number of sums to be able to prepare themselves for the final paper. In this article, we will cover the arithmetic mean, its properties and most importantly, its use in real life.

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Arithmetic mean in simple words is often referred to as average and mean. The simplest way to calculate the mean is by adding all the data and dividing it by the total number of data. It saves a lot of time and further assures accuracy. There are different approaches that can be used to calculate arithmetic mean and students need to gain the knowledge of when to use which approach.

Mathematically, it is equal to the ratio of the sum of numbers in a given set to the total number of values present in the set. In other words, to find the mean of a set of data, add up all the values and then divide this total by the number of values. The mathematical symbol or notation for average is \(\overline x ,\) read as \(x{\rm{ bar}}.\)

The mean of \(n\) observations (variables) \(\;{x_{1,\;\;\;}}{x_{2,\;\;\;}}{x_{3,\;\;\;}}{x_{4,\;}}…..{x_n}\) is given by the formula:

Mean \(\frac{{{x_{1\; + \;}}{x_{2\; + \;\;}}{x_{3\; + \;….\;}}{x_{n\;\;}}}}{n} = \frac{{\sum {x_i}}}{n}\,\) where \(\sum {x_i} = {x_{1 + \;}}{x_{2\; + \;\;}}{x_{3\; + \;\;\;}}{x_{4\; + \;}}…..{x_n}.\) Thus, mean \({\rm{ = }}\frac{{{\rm{Sum\;of\;all\;observations}}}}{{{\rm{Total\;number\;of\;observations}}}}\)

The Greek letter \(\sum\) represents the sum

The arithmetic mean for a grouped data can be obtained through direct method. The formula to find the arithmetic mean with the help of direct method is as follows:

Let \({x_{1,\;\;\;}}{x_{2,\;\;\;}}{x_{3,\;\;\;}}{x_{4,\;}}…..{x_n}\) be the observations with the frequency \({f_{1,\;\;\;}}{f_{2,\;\;\;}}{f_{3,\;\;\;}}{f_{4,\;}}…..{f_n}.\)

Then, the mean is calculated using the formula:

\(\bar x = \frac{{{x_{1\;}}{f_{1\; + \;{x_{2\;}}{f_{2\; + \;}}{x_{3\;}}{f_{3\; + \;}}{x_{4\;}}{f_{4\; + \;\;\;}}……{x_{n\;}}{f_{n\;\;\;}}\;}}}}{{\sum {f_{i\;\;\;}}}}\)

Here, \({f_1} + \;{f_2}\; + \;{f_3}\; + \;\;{f_4}\; + ….\;{f_n} = \sum {f_i}\) indicates the sum of all frequencies.

In generalised form, we can write the arithmetic mean direct method formula as, \(\bar x = \;\frac{{\sum {x_i}{f_i}}}{{\sum {f_i}}}\)

In the assumed mean method, students need to first assume a certain number within the data as the mean. Then, we will calculate the deviation of different classes from the assumed mean, and we will calculate the weighted average of the deviations with the weights being the frequencies and the average is added to the assumed mean.

Thus, the mean is calculated by the formula: \(\overline x = a + \frac{{\sum {f_i}{d_i}}}{{\sum {f_i}}}\)

Where, \(a\) is the assumed mean, \({{f_i}}\) denoted the frequency of \({i^{th}}\) class which is having the deviation of \({d_i}\) from the assumed mean.

**Arithmetic Progression (AP): **A sequence is said to be in arithmetic progression if the difference between a term and its previous term is always the same. In simple terms, it is a sequence where the differences between every two consecutive terms are the same, i.e., \({a_{n + 1}} – {a_{n\;}} = \;d,\) where \({a_{n\;}}\) is the first term, \({a_{n + 1}}\) is the next consecutive term and \(d\) is the difference.

Consider any two numbers, let’s say \(a\) and \(b.\)

And \(P\) be the arithmetic mean between two numbers.

Then the sequence will be as follows: \(a,\;P,\;b\) in A.P.

\(P\; – a\; = \;b\; – \;P\)

\(P\; = \;\frac{{\left( {{\rm{Sum\;of\;the\;numbers}}} \right)}}{{\left( {{\rm{number\;of\;terms}}} \right)}}\)

\(P\; = \;\frac{{\left( {a + b} \right)}}{2}.\)

Now, if \(n\) arithmetic numbers are to be inserted between \(a\) and\(b,\) then we first find the common difference \(d\) which will make the sequence as arithmetic progression.

Here, \(d = \frac{{b – a}}{{n + 1}}.\)

So, the first arithmetic mean will be \(a + d = a + \frac{{b – a}}{{n + 1}},\)Second arithmetic mean will be \(a + 2d = a + 2\frac{{b – a}}{{n + 1}}\) and so on.

We can calculate the arithmetic mean (AM) in three different types of series as listed below.

**Individual series:**Individual series is the series in which each of the units is listed singly as they are observed.

If \({x_1},{x_2},{x_3},{x_4}, \ldots {x_n}\) are the \(n\) items, then the AM can be calculated as,

\(\bar x = \frac{{{x_1} + {x_2} + {x_3} + {x_4} + \;….. + {x_n}}}{n} = \frac{\sum x_i}n\)**Discrete series:**Discrete series is the series where frequencies of a variable are given but the variable is without a class interval.

If \({x_1},{x_2},{x_3},{x_4} \ldots ..{x_n}\) are the \(n\) items and \({f_1},\;{f_2},\;{f_3},\;{f_4} \ldots .{f_n}\) are the corresponding frequencies, then the mean can be calculated as,

\(\bar x = \frac{{{f_1}{x_1} + {f_2}{x_2} + {f_3}{x_3}….. + {f_n}{x_n}}}{n}\; = \;\frac{{\sum x_i f_i}}{n}\)

Where \(n = \sum f.\)**Continuous series:**Those series where frequencies are given along with the value of the variable in the form of class intervals. The mid-point is calculated as

\(\frac{{{\rm{lower\;limit + \;upper\;limit\;}}}}{{\rm{2}}}\)

Thus, the formula to compute mean in continuous is given below;

\(\bar x = \frac{{{f_1}{x_1} + {f_2}{x_2} + {f_3}{x_3} \ldots .. + {f_n}{x_n}}}{n}\)

Where , \(n\) is the number of observations, \({f_{1,\;\;\;}}{f_{2,\;\;\;}}{f_{3,\;}} \ldots .{f_n}\) are the different values of frequency and \({x_{1,\;\;\;}}{x_{2,\;\;\;}}{x_{3,\;}} \ldots .{x_n}\) are the different values of midpoints for ranges.

Here we will familiarise ourselves with some important properties of the arithmetic mean.

If \(\bar x\) is the arithmetic mean of \(n\) observations, \({x_{1,\;\;\;}}{x_{2,\;\;\;}}{x_{3,\;\;\;}}{x_{4,\;\;}}…..{x_n}\;,\) then

\(\left( {{x_1}\; – \;\bar x} \right)\; + \;\left( {{x_{2\;}} – \;\bar x} \right)\; + \;\left( {{x_{3\;}} – \;\bar x} \right)\; + \;…\; + \;\left( {{x_n}\; – \;\bar x} \right)\; = \;0.\)

- The mean of \(n\) observations \({x_{1,\;\;\;}}{x_{2,\;\;\;}}{x_{3,\;\;\;}}{x_{4,\;\;}}…..{x_n}\;\) is \(\overline x .\) If each observation is increased by \(y,\) the mean of the new observations is \(\left( {\bar x\; + \;y} \right).\)
- The mean of \(n\) observations \({x_{1,\;\;\;}}{x_{2,\;\;\;}}{x_{3,\;\;\;}}{x_{4,\;\;}}…..{x_n}\) is \(\overline x .\) If each observation is decreased by \(y,\) the mean of the new observations is \(\left( {\bar x\; – \;y} \right).\)
- The mean of n observations \({x_{1,\;\;\;}}{x_{2,\;\;\;}}{x_{3,\;\;\;}}{x_{4,\;\;}}…..{x_n}\) is \(\bar x.\) If each observation is multiplied by a non-zero number \(y,\) the mean of the new observations is \(\left( {y \times \bar x\;} \right).\)
- The mean of n observations \({x_{1,\;\;\;}}{x_{2,\;\;\;}}{x_{3,\;\;\;}}{x_{4,\;\;}}…..{x_n}\) is \(\bar x.\) If each observation is divided by a non-zero number \(y,\) the mean of the new observations is \(\left( {\frac{{\bar x}}{y}} \right).\)
- If all the observations in the given data set have a value say \(‘y’,\) then their arithmetic mean is also \(‘y’.\)

There are always pros and cons whenever we talk about anything. Arithmetic mean too, has its merits and demerits.

Let’s talk first about the advantages of the arithmetic mean.

- As the formula to find the arithmetic mean is rigid, the result does not change.
- It takes into consideration each value of the data set.
- Finding the arithmetic mean is quite simple.
- It’s also a useful measure of central tendency, as it tends to provide useful results, even with a large group of numbers.
- It can be further subjected to many algebraic treatments, unlike mode and median. For example, the mean of two or more series can be obtained from the mean of the individual series.
- The arithmetic mean is widely used in geometry as well. For example, the coordinates of the “centroid” of a triangle (or any other figure bounded by line segments) are the arithmetic mean of the coordinates of the vertices.

**Disadvantages of Arithmetic Mean**

- The arithmetic mean is affected by extreme values in the data set.
- In a distribution of open-end classes, the value of the mean cannot be computed without making assumptions regarding the size of the class.
- It’s practically impossible to locate the arithmetic mean by inspection or graphically.
- We can’t find the arithmetic mean if a single observation is missing or lost.

** Q.1. Find the mean of first **\(\mathbf6\)

Sum of these multiples \( = 5 + 10 + 15 + 20 + 25 + 30 = 105\)

Number of multiples \(= 6\)

Average \({\rm{ = }}\frac{{{\rm{Sum\;of\;6\;multiples}}}}{{{\rm{Number\;of\;multiples}}}}\)

Average \({\rm{ = }}\frac{{5 + 10 + 15 + 20 + 25 + 30}}{6}\)

Average \( = \frac{{105}}{6} = 17.5\)

Hence, the arithmetic mean of the first \(6\) multiples of \(5\) is equal to \(17.5.\)

**Q.2. If the mean of **\(\boldsymbol y\boldsymbol+\mathbf2\boldsymbol,\boldsymbol\;\boldsymbol y\boldsymbol+\mathbf4\boldsymbol,\boldsymbol\;\boldsymbol y\boldsymbol+\mathbf6\boldsymbol,\boldsymbol\;\boldsymbol y\boldsymbol+\mathbf8\boldsymbol\;\boldsymbol a\boldsymbol n\boldsymbol d\boldsymbol\;\boldsymbol y\boldsymbol+\mathbf{10}\)

Mean \(= \;\frac{{{\rm{Sum\;of\;numbers}}}}{{{\rm{Number\;of\;numbers}}}}\)

\(13\; = \;\frac{{y + 2 + y + 4 + y + 6 + y + 8 + y + 10}}{5}\)

On further calculation, we get

\(13\; \times \;5\; = \;5y\; + \;30\)

\(65\; = \;5y\; + \;30\)

\(5y\; = \;65\;–\;30\)

\(5y\; = \;35\)

\(y\; = \;7.\)

*Q.3*. *Find the mean of the following distribution:*

\(x\) | \(10\) | \(30\) | \(50\) | \(70\) | \(90\) |

\(f\) | \(5\) | \(6\) | \(5\) | \(10\) | \(4\) |

**Answer**:

\({x_i}\) | \({f_i}\) | \({f_i}{x_i}\) |

\(10\) | \(5\) | \(10 \times 5 = 50\) |

\(30\) | \(6\) | \(30 \times 6 = 180\) |

\(50\) | \(5\) | \(50 \times 5 = 250\) |

\(70\) | \(10\) | \(70 \times 10 = 700\) |

\(90\) | \(4\) | \(90 \times 4 = 360\) |

Total | \(\sum {f_i} = 30\) | \(\sum {x_i}{f_i} = 1540\) |

Now, the mean formula is \(\bar x\; = \;\frac{{\sum {x_i}{f_i}}}{{\sum {f_i}}}\) \(\Rightarrow \bar x = \;\frac{{1540}}{{30}}\) \(\Rightarrow \bar x = 51.33\) Hence, the required mean is \(51.33.\)

*Q.4.* *Find the mean of the following distribution:*

Class-Interval | \(15-25\) | \(25-35\) | \(35-45\) | \(45-55\) | \(55-65\) | \(65-75\) | \(75-85\) |

Frequency | \(6\) | \(7\) | \(6\) | \(4\) | \(4\) | \(2\) | \(1\) |

** Ans**: When the data is presented in the form of class intervals, we must find the mid-point of each class, which is known as the Class Mark.

Class Mark \( = \frac{{{\text{ Upper limit }} + {\text{ Lower limit }}}}{2}\)

Class-Interval | Class Mark \(\left( {{x_i}} \right)\) | Frequency \(\left( {{f_i}} \right)\) | \({x_i}{f_i}\) |

\(15-25\) | \(20\) | \(6\) | \(120\) |

\(25-35\) | \(30\) | \(7\) | \(210\) |

\(35-45\) | \(40\) | \(6\) | \(240\) |

\(45-55\) | \(50\) | \(4\) | \(200\) |

\(55-65\) | \(60\) | \(4\) | \(240\) |

\(65-75\) | \(70\) | \(2\) | \(140\) |

\(75-85\) | \(80\) | \(1\) | \(80\) |

Total | \(30\) | \(1230\) |

Now, the mean formula is \(\bar x\; = \;\frac{{\sum {x_i}{f_i}}}{{\sum {f_i}}}\) \(\Rightarrow \bar x = \;\frac{{1230}}{{30}}\) \(\Rightarrow \bar x = 41\) Hence, the required mean is \(41.\)

**Q.5. In an examination, the mean of marks scored by a class of \(40\) students was calculated as \(72.5\). Later on, it was detected that the marks of one student were wrongly copied as \(48\) instead of \(84.\) Find the correct mean.**

\(72.5 = \frac{{{\rm{Incorrect\;sum\;of\;marks\;of}}\;40\;{\rm{students}}\;}}{{40}}\)

Incorrect sum of marks of

Since the marks of one student were wrongly copied as \(48\) instead of \(84,\)

Correct sum of marks of \(40\) students \( = 2900 – 48 + 84 = 2936.\)

Correct mean \( = \frac{{2936}}{{40}} = 73.4.\)

In this article, we learned about the arithmetic mean, the need for it and its formula. We also learned about the properties and the advantages and disadvantages of the arithmetic mean. In addition to that, we also learned to find the arithmetic mean formula for grouped data and ungrouped data.

Frequently asked question related to arithmetic mean is listed as follows:

*Q. What is arithmetic mean?***A. **For beginners arithmetic mean is the same as average. There are a variety of data available and considering the data type, students need to decide the correct approach that is appropriate for the concerned data.

*Q. Is arithmetic mean an important chapter from the exam perspective?***A. **It is advisable for students to cover all the topics included in the syllabus and not engage in any kind of selective studies.

*Q. Where can I find the PDF of arithmetic mean?***A. **You can find the PDF of arithmetic mean on this page, students can download it for free.

*Q.**Which approach of arithmetic mean is important?***A. **All the approaches related to finding arithmetic mean is important. Students need to practice to be able to identify the correct approach considering the data type.

*Q. Where can I find previous year question papers?***A. **Students can find previous year question papers on this page itself. It is necessary to go through the previous year question papers to be able to improve their performance.

Now that you are provided with all the necessary information on the arithmetic mean and its formulas, we hope this article is helpful to you. If you have any queries on this page, post your comments in the comment box below and we will get back to you as soon as possible.

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