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9th Class Mathematics Statistics Mean

Mean

Category : 9th Class

*       Mean

 

It is also known as arithmetic mean of a given observation is equal to ratio of sum of observation and total number of observation, i.e.

\[\text{Mean }=\frac{\text{Sum of total observation}}{\text{Total Number of observation}}\]

If \[{{x}_{1}},{{x}_{2}},.....{{x}_{n}}\] are n observation then its mean                

\[\overline{\text{X}}=\frac{{{x}_{1}}+{{x}_{2}}+.....+{{x}_{n}}}{n}=\frac{\sum {{x}_{i}}}{n}\]

The arithmetic mean of grouped data calculated by the following methods:

(i) Direct method

(ii) Assumed mean method

(iii) Step - Deviation method  

 

*            Direct Method

In this method, suppose \[{{x}_{1}},{{x}_{2}},.....,{{x}_{n}}\] are the observation having frequency\[{{f}_{1}},{{f}_{2}},.....,{{f}_{n}}\] respectively then

\[\overline{\text{X}}=\frac{{{x}_{1}}{{f}_{1}}+{{x}_{2}}{{f}_{2}}+.....,+{{x}_{n}}{{f}_{n}}}{{{f}_{1}}+{{f}_{2}}+.....{{f}_{n}}}=\frac{\sum {{x}_{i}}{{f}_{i}}}{\sum {{f}_{i}}}\]

 

Direct method for calculating mean depend on the following steps:

Step 1: Find the class mark (which is discussed in previous class) for each class.

Step 2: Calculate product of frequency and class mark for each class interval.

Step 3: Then calculate mean by using formula \[\left( \frac{\sum fixi}{\sum fi} \right)\].  

 

 

 

Find the mean of the following data:

Class Interval Frequency
0 - 10 16
10 - 20 12
20 - 30 9
30 - 40 6
40 - 50 7

(a) 20.2                                

(b) 21.3          

(c) 45.5                                

(d) 43.2                                

(e) None of these

 

Answer: (a)  

Explanation

Class Interval Frequency (\[{{f}_{i}}\]) Class Mark \[({{x}_{i}})\] \[{{f}_{i}}{{x}_{i}}\]
0 - 10 16 5 80
10 - 20 12 15 180
20 - 30 9 25 225
30 - 40 6 35 210
40 - 50 7 45 315
  \[\sum {{f}_{i}}=50\]   \[\sum {{f}_{i}}{{x}_{i}}=1010\]

 

\[\therefore \] Mean\[=\frac{1010}{50}=20.2\]  

 

*            Assumed Method for Calculating Mean

For calculating mean, we should follow the following steps:

Step 1: For each class interval, calculate the class mark using the formula\[{{x}_{i}}=\frac{1}{2}\] (lower limit + upper limit)

Step 2:   Choose a suitable value of\[x\]. in the middle as the assumed mean and denoted by A.

Step 3:   Calculate the deviations\[{{d}_{i}}({{x}_{i}}-A)\]for each\[i\].

Step 4:   Calculate the product\[{{f}_{i}}{{d}_{i}}\]for each\[i\].

Step 5:   Calculate summation of frequency.

Step 6:   Then at last calculate mean, by using formula\[\overline{X}=A+\frac{\sum {{f}_{i}}{{d}_{i}}}{n}\] when \[n=\sum {{f}_{i}}\]  

 

 

Find the mean of the following data:

Class Interval Frequency
0 - 10 7
10 - 20 8
20 - 30 12
30 - 40 13
40 - 50 10

 

(a) 25.6                

(b) 27.2                

(c) 23.3                                

(d) 24.5                                

(e) None of these

 

Answer: (b)

Explanation

Class Interval Frequency (\[{{f}_{i}}\]) Class Mark \[({{x}_{i}})\] Deviation \[{{f}_{i}}{{d}_{i}}\]
0 - 10 7 5 -20 -140
10 - 20 8 15 -10 -80
20 - 30 12 25=A 0 0
30 - 40 13 35 10 130
40 - 50 10 45 20 200
  \[\sum {{f}_{i}}=50\]     \[\sum {{f}_{i}}{{d}_{i}}=110\]

 

If A=25

Mean\[=\overline{X}=A+\frac{\sum {{f}_{i}}{{d}_{i}}}{n}\]

\[25+\frac{110}{50}=27.2\]  

 

*            Step - Deviation Method

If the value of class marks and frequency are large then calculating mean by above method is very difficult, due to this region this method can be used. This method depends on the following steps:

Step 1: For each class interval, calculating the class marks\[{{x}_{i}}\]. where\[{{x}_{i}}=\frac{1}{2}\] (lower limit + upper limit).

Step 2: Choose a suitable value of\[x\], in the middle as the x; column as the assumed mean and denoted it by A.

Step 3: Calculate h = (upper limit) - (lower limit), which is the same for all the classes.

Step 4: Calculate\[{{u}_{i}}=\frac{{{x}_{i}}-A}{h}\]for each class.

Step 5: Calculate for each class and hence, find\[\sum ({{f}_{i}}\times {{u}_{i}})\].

Step 6: Calculate the mean by using the formula\[\overline{X}=A+\left[ h\times \frac{\sum ({{f}_{i}}\times {{u}_{i}})}{\sum {{f}_{i}}} \right]\].

 

 

Find the mean of the following frequency distribution:

(a) 112.2                                              

(b) 113.2         

(c) 114.5                                              

(d) 115.3

(e) None of these  

 

Answer: (a)  

Explanation  

Class Interval Frequency (\[{{f}_{i}}\]) Class Mark \[({{x}_{i}})\] \[{{u}_{i}}=({{x}_{i}}-A)/n\] \[{{f}_{i}}{{u}_{i}}\]
50 - 70 18 60 -2 -36
70 - 90 12 80 -1 -12
90 - 110 13 100=A 0 0
110 - 130 27 120 2 37
130 - 150 8 140 2 16
150 - 170 22 160 3 66
  \[\sum {{f}_{i}}=100\]     \[\sum {{f}_{i}}{{u}_{i}}=110\]

 

From table, A = 100, h = 20, \[\sum {{f}_{i}}=100\]and\[\sum {{f}_{i}}\times {{u}_{i}}=61\]  

Mean, \[\overline{X}=A+\left[ h\times \frac{\sum ({{f}_{i}}\times {{u}_{i}})}{\sum {{f}_{i}}} \right]=100+\left[ 20\times \frac{61}{100} \right]=112.2\]

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