x class statistics-arithmetic mean

Statistics is the science of collection, organization, presentation, analysis, and interpretation of numerical data.

Measures of central tendency

A single value that describes the characteristics of the entire data is known as the measure of central tendency. The following are commonly used measures of central tendency.

(i) Mean (or)Average (or) Arithmetic Mean (AM)

(ii) Median

(iii) Mode.

Arithmetic Mean:

The mean of observations is the sum of the values of all the observations divided by the total number of observations. It is denoted by the symbol `\bar{x}`, read as x bar.

Mean of Ungrouped data:

Let x₁,x₂,. . ., x_n are n observations then

Example: Find the mean of the first 5 natural numbers.

Soution:

Mean = `(1+2+3+4+5)/5 = 15/5 = 3`

Example: Find the mean of factors of 24.

Solution:

Factors of 24 is 1,2,3,4,6,8,12,24.

`\therefore`Mean=`(1+2+3+4+6+8+12+24)/8= 60/8 = 7.5`

Example: If the mean of 5,7,9,p is 9 then the value of p is.

Solution:

Mean = `\frac{Sum  of  the  values  of  all  the  observations}{Total  Number  of  observations}`

`\therefore`9=`(5+7+9+p)/4= (21+p)/4 `

21+p=36 ⇒ p=36-21=15

properties of mean :

• If a constant real number ‘a’ is added to each of the observations then the new mean will be `\bar{x}` + a.
• If a constant real number ‘a’ is subtracted from each of the observations then-new mean will be `\bar{x}` − a
• If a constant real number ‘a’ is multiplied with each of the observations then-new mean will be a`\bar{x}`
• If each of the observations is dived by a constant no ‘a’ then-new mean will be `\frac{\bar{x}}{a}`.

Mean of Grouped data:

Example-1:

The marks obtained in mathematics by 30 students of Class X of a certain school are given in the table below. Find the mean of the marks obtained by the students.

Solution:

To find the mean marks we require the product of each x_i with the corresponding frequency f_i.

` \bar x = \frac{sum f_{i}x_{i}}{sum f_{i}} = 1779/30 = 59.3`

`\therefore`The mean marks obtained is 59.3.

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Direct Method:

Here f_i= frequency of i^thclass

x_i= midpoint of i^thclass (classmark)

Class mark = `(Upper  class  liMit + Lower  class  liMit)/2`

for example in the class 10- 20 class mark = `(10+20)/2 = 30/2=15`

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Example: Find the mean for the folowing distribution

Solution:

Marks	Mid value(xi)	Frequency(fi)	fixi
10-20	15	6	90
20-30	25	8	200
30-40	35	13	455
40-50	45	7	345
50-60	55	3	165
60-70	65	2	130
70-80	75	1	75
Total		`sum f_i =40`	`sum f_{i}x_{i}=1430`

` \bar x = \frac{sum f_{i}x_{i}}{sum f_{i}} = 1430/40 = 35.75`

Example: Find the mean for the folowing distribution

Solution:

` \bar x = \frac{sum f_{i}x_{i}}{sum f_{i}} = 1860/30 = 62`

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Deviation Method (or) Assumed Mean Method

Sometimes when the numerical values of x_i and f_i are large, finding the product of x_i and f_i becomes large and time-consuming. So, for such situations, let us think of a method of reducing these calculations. We can do nothing with the f_i ’s, but we can change each x_i to a smaller number so that our calculations become easy.

(i)For that we choose one among the x_i’s as the assumed mean, and denote it by ‘a’.

(ii)Then find the difference d_i between a and each of the x_i’s

So d_i= x_i - a

(iii)Then find the product of d_i with the corresponding f_i, and take the sum of all the f_id_i’s.

Example: Find the mean for the following distribution with the Assumed mean method

Solution:

Let us take Assumed mean (a) = 47.5

` \bar x = a+ \frac{sum f_{i}d_{i}}{sum f_{i}} `

` \bar x = 47.5+ 435/30 = 47.5+14.5 = 62`

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Step-deviation Method:

Here

a= assumed mean

h=class size

`u_i=(x_i-a)/h`

f_i= frequency of i^thclass

x_i= midpoint of i^thclass (class mark)

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Example: Find the mean for the following distribution using the step deviation method

Solution:

Let we take Assumed mean (a) = 47.5

` \bar x = a+ h\frac{sum f_{i}u_{i}}{sum f_{i}} `

` \bar x = 47.5+ 15(29/30) = 47.5+14.5 = 62`

So, the mean marks obtained by a student is 62.

Example2:

Find the mean percentage of female teachers by all the three methods from the below-given table

Solution:

Let we take Assumed mean (a) = 50

`sum f_i=35 `

`sum f_{i}x_i = 1390`

`sum f_{i}d_i =-360`

` sum f_{i}u_i = -36`

Using direct method `\bar x = \frac{sum f_{i}x_{i}}{sum f_{i}} `

`\bar x=1390/35=39.71`

` \bar x = a+ \frac{sum f_{i}d_{i}}{sum f_{i}} `

`\bar x=50 +(-360)/35=50-10.29=39.71`

` \bar x = a+ h\frac{sum f_{i}u_{i}}{sum f_{i}} `

` \bar x = 50+ 10((-36)/35) = 50-10.29 = 39.71`

`\therefore` The mean percentage of female teachers in the primary schools of rural areas is 39.71.

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Note:

• the step-deviation method will be convenient to apply if all the di’s have a common factor.
• The mean obtained by all three methods is the same.
• The assumed mean method and step-deviation method are just simplified forms of the direct method.

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Use of Arithmetic Mean

• It is widely used in obtaining practical statistics and estimates.
• It is used to calculate the average marks obtained by the student.
• Traders use it to find profit per unit article, output per machine, average monthly profit Income and expenditure etc.

Merits of Arithmetic Mean

• Mathematical analysis of the mean is possible. So, it is relatively reliable.
• It is strictly defined, simple, easy to understand, and easy to calculate.
• It is based on all observations.
• Its value is unique and we can use it to compare different sets of data.
• It is at least affected by the sample fluctuations.

Demerits of Arithmetic Mean:

• It shall not be determined by inspection or marked graphically.
• It cannot be obtained if a single observation is missed.
• It is very much influenced by extreme values. In the case of serious goods, A.M. Gives a distorted image of Distribution and is no longer representative of distribution.
• It may lead to erroneous conclusions if details of the calculated data are not given.

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Statistics formula sheet

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