Properties of Arithmetic Mean | Merits and Demerits of Arithmetic Mean | AM Properties with Proof

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Arithmetic Mean(AM) is the most important concept in statistics. It is one of the measures of central tendency that can be directly described as the sum of all quantities to be divided by the number of quantities. Every time we can’t apply the formula of AM to solve the problems on average or mean or arithmetic mean. So, we have explained the properties of arithmetic mean with proofs which aid students to calculate different types of questions on average with ease.

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What are the Properties of Arithmetic Mean?

Properties of AM are used to solve complex problems based on mean/arithmetic mean/average. Some of the important arithmetic mean properties that are used in solving the problems based on average are mentioned here briefly. Just take a look at them and be aware of all properties to use.

  • If \(\overline{x}\) is the arithmetic mean of n observations, x1, x2, x3, …., xn, then (x1–\(\overline{x}\))+(x2–\(\overline{x}\))+(x3–\(\overline{x}\))+…+(xn–\(\overline{x}\))=0.
  • The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is increased by y, the mean of the new observations is (\(\overline{x}\)+y).
  • The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is decreased by y, the mean of the new observations is (\(\overline{x}\)–y).
  • The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is divided by a non-zero number y, the mean of the new observations is (\(\overline{x}\)*y).
  • The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is multiplied by a non-zero number y, the mean of the new observations is (y*\(\overline{x}\)).
  • If all the observations in the given data set have a value say ‘y′, then their arithmetic mean is also ‘y′.

Illustration of Arithmetic Mean Properties

Property 1: 

If \(\overline{x}\) is the arithmetic mean of n observations, x1, x2, x3, …., xn, then (x1–\(\overline{x}\))+(x2–\(\overline{x}\))+(x3–\(\overline{x}\))+…+(xn–\(\overline{x}\))=0.

Proof:

We know that

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Hence, (x1–\(\overline{x}\))+(x2–\(\overline{x}\))+(x3–\(\overline{x}\))+…+(xn–\(\overline{x}\))

= (x1 + x2 + x3 + . . . + xn) – n\(\overline{x}\)

= (n\(\overline{x}\) – n\(\overline{x}\)), [using (i)].

= 0.

Therefore, (x1–\(\overline{x}\))+(x2–\(\overline{x}\))+(x3–\(\overline{x}\))+…+(xn–\(\overline{x}\))=0.

Property 2:

The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is increased by A, the mean of the new observations is (\(\overline{x}\) + A).

Proof: 

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Mean of (x1 + A), (x2 + A), …, (xn + A)

= {(x1 + A) + (x2 + A) + … + (xn+ A)}/n

= {(x1 + x2 + . . . + xn) + nA}/n

= (n\(\overline{x}\) + nA)/n, [using (i)].

= {n(\(\overline{x}\) + A)}/n

= (\(\overline{x}\) + A).

So, the mean of the new observations is (\(\overline{x}\) + A).

Property 3: 

The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is decreased by p, the mean of the new observations is (\(\overline{x}\) – a).

Proof:

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Mean of (x1 – p), (x2 – p), …., (xn – p)

= {(x1 – p) + (x2 – p) + … + (xn – p)}/n

= {(x1 + x2 + . . . + xn) – np}/n

= (n\(\overline{x}\) – np)/n, [using (i)].

= {n(\(\overline{x}\) – p)}/n

= (\(\overline{x}\) – p).

Hence, the mean of the new observations is (\(\overline{x}\) + p).

Property 4:

The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is multiplied by a nonzero number p, the mean of the new observations is p\(\overline{x}\).

Proof:

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Mean of px1, px2, . . ., pxn,

= (px1 + px2 + … + pxn)/n

= {p(x1 + x2 + . . . + xn)}/n

= {p(n\(\overline{x}\))}/n, [using (i)].

= p\(\overline{x}\).

Hence, the mean of the new observations is p\(\overline{x}\).

Property 5:

The mean of n observations x1, x2, x3, …., xn is \(\overline{x}\). If each observation is divided by a nonzero number p, the mean of the new observations is (\(\overline{x}\)/p).

Proof:

\(\overline{x}\) = (x1 + x2 + x3 + . . . + xn)/n

⇒ (x1 + x2 + x3 + . . . + xn) = n\(\overline{x}\). ………………….. (i)

Mean of (x1/p), (x2/p), . . ., (xn/p)

= (1/n) ∙ (x1/p + x2/p + …. + xn/p)

= (x1 + x2 + . . . + xn)/np

= (n\(\overline{x}\))/(np), [using (i)].

= (\(\overline{x}\)/p).

Merits of Arithmetic Mean

  1. It is determined strictly.
  2. It is calculated on the basis of all observations
  3. Simple to solve and easy to understand
  4. It is responsive to mathematical treatment or properties.
  5. It is inspired by the value of every item in the observation series.

Demerits of Arithmetic Mean

  1. If any single observation is missing or lost then it’s unable to find the arithmetic mean of the data.
  2. By inspection or graphically, it is quite difficult to find the arithmetic mean.
  3. By the extreme values in the set of the data, the AM gets affected.
  4. In some situations, the arithmetic mean does not exemplify the original item.

Solved Problems on Mathematical Properties of Arithmetic Mean Proof

Example 1:
If the two variables x and y are related by 3x + 4y + 6 = 0 and x̄ = 10, then Arithmetic mean of “y” = (-6 – 3x̄) / 6, Find AM of y?

Solution:
Given x̄ = 10
Arithmetic mean of “y” = (-6 – 3x̄) / 6
Arithmetic mean of “y” = (-6 – 3×10) / 6
Arithmetic mean of “y” = (-6 – 30) / 6
Arithmetic mean of “y” = (-36) / 6
Arithmetic mean of “y” = -6

Example 2:
If a variable “x” assumes 7 observations, say 11, 22, 33, 44, 55, 66, 77 then x̄ = 44. Calculate the instance using property 1.
Solution:
Given 7 observations are 11, 22, 33, 44, 55, 66, 77 and it’s mean is 44
Now, by using property 1 ie., ∑(x – x̄) = 0.
The deviations of the observations from the arithmetic mean (x – x̄) are -33, -22, -11, 0, 11, 22, 33
Now, ∑(x – x̄) = -(33) + (-22) + (-11) + 0 +11 + 22 + 33 = 0.

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