如何找到算术平均值？

算术可能在当时拥有最长的历史。它是一种计算方法，自古以来就用于常规计算，如测量、标记和各种日常计算，以获得确定的值。该术语起源于希腊语“arithmos”，它的意思是数字。

Arithmetic is the elementary branch of mathematics that specifically deals with the study of numbers and properties of traditional operations like addition, subtraction, multiplication, and division.

除了传统的加法、减法、乘法和除法运算外，还包括百分比、对数、指数和平方根等高级计算。算术是数学的一个分支，涉及数字及其传统运算。

算术史

算术是数学领域中最长的。它自人类文明的远古时代就在使用。它最初用于查找对象的总和和差异以及用于货币化目的。

• The 17th-century Indian mathematician Brahmagupta is the “father of arithmetic”
• Carl Friedrich Gauss in 1801, provided the Fundamental principle of number theory.

什么是算术平均值？

Arithmetic mean is defined as the sum of all of the numbers of the group, when divided by the number of items present in it.

算术平均值也简称为平均值或平均值。用于算术平均值的符号是“X”。它是一种计算集中趋势的方法。可以简单理解为对整个数据进行计算得到的单个平均值。

算术在日常计算中很有用，因为它给出了一个确定的值，并且由于它基于观察而易于计算和理解。然而，因为也有一些案例给出了一个荒谬的价值，这被认为是一个过失。

算术平均值的公式由下式给出：

${\displaystyle A.M={\frac {1}{n}}\sum _{i=1}^{n}a_{i}={\frac {a_{1}+a_{2}+\cdots +a_{n}}{n}}}$

Where,

• N is the number of items
• A.M is the arithmetic mean
• a_i  is set values

现在，让我们进入这个问题。

如何找到算术平均值？

我们可以通过将组的所有数字相加并将总和除以项目数来计算算术平均值。

它的数学表示可以表示为-

X = Sum of Values/N

Where 

X is the arithmetic mean 

N is the number of items

例如，数字 2、3、4 和 7 的算术平均值将由下式给出

X = (2 + 3 + 4 + 7)/4

= 16/4 = 4

示例问题

问题1：班上4名学生的数学成绩分别为10、12、22、30，求算术平均值。

解决方案：

As we know,

A.M = Sum of all numbers/Total  no. of items

A.M = (10 + 12 + 22 + 28)/4

= 72/4 = 51

Therefore, the mean is 51.

问题 2：如果 5 个观测值 6、7、5、x、10 的算术平均值为 6。求 x 的值。

解决方案：

Given,

The 5 observations are 6, 7, 5, x, 10

no. of items (n) = 5

Arithmetic mean = 6

We have,

X = Sum of all numbers/Total no. of items

=>6 = (6 + 7 + 5 + x + 10)/5

=>30 = 28 + x

=>x = 30 – 28 = 2

Therefore, the value of missing observation is 2.

问题3：如果2a+3、a+2、3a+4、4a+5的算术平均值为a+2，求a的值。

解决方案：

Given,

The observations are 2a+3, a+2, 3a+4 and 4a+5.

hence, no. of items (n) = 4

Arithmetic mean (X) = a + 2

We have, 

X = Sum of all numbers/Total no. of items

=>X = [(2a + 3) + (a + 2) + (3a + 4) + (4a + 5)/4

=>a + 2 = (10a + 14)/4

=>4(a + 2) = (10a + 14)

=>4a + 8 = 10a + 14

=>(-6a) = (-6)

=>a = -1

Therefore, the value of a is -1.

问题 4：Subham 在期末考试中获得的分数分别为 50、35、40、45 和 55。计算算术平均值。

解决方案：

Given,

The observations are 50, 35, 40,45, and 55.

No. of items (n) = 5

Arithmetic mean(X) = ?

We have,

X = Sum of all numbers/ Total no. of items

=>X = (50 + 35 + 40 + 45 + 55)/5

=>X = 225/5

=>X = 225/5 = 45

Therefore, the arithmetic mean of marks obtained is 45.

问题 5：求前五个素数的平均值。

解决方案：

The first five prime numbers are 2, 3, 5, 7 and 11.

No. of observations (n) = 5

We have,

X = Sum of all numbers/ No. of items

=>X = (2 + 3 + 5 + 7 + 11)/5

=>X = 28/5 = 5.6

Therefore, the arithmetic mean of the first five prime numbers is 5.6.

问题6：5个观测值m、m+4、m+6、m+8、m+12的算术平均值为16。求m的值。

解决方案：

Given, 

The observations are m, m+4, m+6, m+8 and m+12.

No. of items (n) = 5

Arithmetic mean (X) = 16

We have, 

X=sum of all numbers/no. of items

=>X = [m + (m + 4) + (m + 6) + (m + 8) + (m + 12)]/5

=>16 = (5m + 30)/5

=>80 = 5m + 30

=>5a = 50

=>m = 10

Therefore, the value of m is 10.