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Arithmetic Mean in statistics is used to find central tendencies of skewed distribution. However, apart from being used in mathematics, its application is evident in almost every academic field like economics, anthropology and history. It is also applicable in finding per capita income of a demographic or nation’s population, measuring average temperature of earth to measure global warming or calculate average score in sports such as cricket.
Key Terms: Arithmetic Mean, average, statics, Direct Method, Short cut method, Deviation Method
What is Arithmetic Mean?
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Arithmetic mean is generally used for calculating a mean or average of a grouped data. It is calculated by dividing the sum of a collection of integers by the number of numbers in the set. It is used when all of the values in a set of data have the same measuring unit, such as heights, miles, hours, and so on.
Consider if you are asked to find the arithmetic mean of 13, 6, 9. Then you first find the sum of the data and divide it with the count of data. 13+ 6+ 8 = 27, so the arithmetic mean is 9. Arithmetic mean is sometimes called just ‘mean’ or ‘average’. In the next section we will see the formula to find the mean of cumulative data.
Formula for Arithmetic Mean
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In a discrete frequency distribution the arithmetic mean may be computed by any one of the following methods:
- Direct Method
- Short cut method
- Step deviation method
Direct Method
If x1, x2, ... , xn has corresponding f1, f2, … fn frequencies, then the arithmetic mean is given by,
X = (f1 x1 + f2 x2 + …. +fn xn) / (f1 + f2 +....+ fn)
Or, X= (\(\sum_{i=1}^{n}\)fi xi) / (\(\sum_{i=1}^{n}\)fi)
Example 1. Find the arithmetic mean of the following distribution:
| x | 4 | 6 | 9 | 10 | 15 |
| f | 5 | 10 | 10 | 7 | 8 |
Solution:
| xi | fi | fi xi |
| 4 6 9 10 15 | 5 10 10 7 8 | 20 60 90 70 120 |
| \(\sum\)fi = 40 | \(\sum\)fi xi = 360 |
Using the formula,
X= (\(\sum_{i=1}^{n}\)fi xi) / (\(\sum_{i=1}^{n}\)fi)
X = 360 / 40 = 90
Therefore the arithmetic mean of the distribution is 90.
Short Cut Method
If values of the frequencies are large then the calculation of arithmetic mean becomes very difficult and time consuming. So, a better approach is the shortcut method.
Let x1, x2, ... ,xn has f1, f2, ... ,fn respective frequencies, taking deviation about a random point ‘A’ we get,
di = xi - A
fi di = fi(xi - A)
\(\sum_{i=1}^{n}\) fi di = \(\sum_{i=1}^{n}\) fi (xi - A)
Or, \(\sum_{i=1}^{n}\)fi di =\(\sum_{i=1}^{n}\) fi xi -\(\sum_{i=1}^{n}\)fi A
Or, \(\sum_{i=1}^{n}\)fi di =\(\sum_{i=1}^{n}\) fi xi -AN [here \(\sum_{i=1}^{n}\)fi = N ]
Or, (1/N)(\(\sum_{i=1}^{n}\)fi di) = (1/N)(\(\sum_{i=1}^{n}\) fi xi) -A
Or, X = A + (1/N)(\(\sum_{i=1}^{n}\) fi di)
Example: Find the Arithmetic mean of the following distribution table:
Weight: 67 70 72 73 75
No. of students: 4 3 2 2 1
Solution:
Let assume A = 72
| xi | fi | di = xi - A xi - 72 | fi di |
| 67 70 72 73 75 | 4 3 3 2 1 | -5 -2 0 1 3 | -20 -6 2 2 3 |
| N= \(\sum\)fi = 12 | \(\sum\)fi di = -21 |
Using the formula, we get
X = A + (1/N)(\(\sum_{i=1}^{n}\)fi di)
X = 72 + (1/12)(-21) = 70.25kg
Therefore the mean weight is 70.25 kg.
Also read Difference Between Variance and Standard Deviation
Step deviation Method
If there is a visible common number in the data distribution then the step deviation method shortens the calculation of the mean.
Consider h is the visible common number
ui = (xi - A)/h
Or, xi = A + h ui
Or, fi xi = fi A +hf ui
Or, \(\sum_{i=1}^{n}\)fi xi =A\(\sum_{i=1}^{n}\) fi + h\(\sum_{i=1}^{n}\)fi ui
Or, (1/N)(\(\sum_{i=1}^{n}\)fi xi) = (A/N)(\(\sum_{i=1}^{n}\) fi) + (h/N) \(\sum_{i=1}^{n}\)fi ui
Or, X = A +h(1/N) \(\sum_{i=1}^{n}\)fi ui
Example: Find the average of the following distribution table:
| Wage | 800 | 820 | 860 | 900 | 920 | 980 | 1000 |
| No. of workers | 7 | 14 | 19 | 25 | 20 | 10 | 5 |
Solution:
Let A = 900 and h = 20
| xi | fi | di = xi - A = xi - 900 | ui = (xi - 900)/20 | fi ui |
| 800 820 860 900 920 980 1000 | 7 14 19 25 20 10 5 | -100 -80 -40 0 20 80 100 | -5 -4 -2 0 1 4 5 | -35 -56 -38 0 20 40 25 |
| \(\sum\)fi = 100 | \(\sum\)fi ui = -44 |
Using the formula, we get,
X = A +h(1/N) \(\sum_{i=1}^{n}\)fi ui
X = 900 + (20/100) (-44) = 900 -8.80 =891.2
Therefore, the average of the given data distribution is 891.2.
Read Also: Degree of Polynomial
Limitations of Arithmetic Mean
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Following are the Arithmetic mean limitations.
- Extreme values have a big impact on the arithmetic mean.
- It is unable to properly average percentages & ratios.
- For strong data distribution, it is not a suitable average.
- If any piece is missing, it is impossible to calculate precisely.
- The mean does not always agree with any of the observations.
Important Topics for JEE MainAs per JEE Main 2024 Session 1, important topics included in the chapter arithmetic mean are as follows:
Some memory based important questions asked in JEE Main 2024 Session 1 include:
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Things to Remember
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- Arithmetic mean is generally used for calculating a mean or average of a grouped data.
- Arithmetic mean is calculated by dividing the sum of a collection of integers by the number of numbers in the set.
- In a discrete frequency distribution the arithmetic mean may be computed by Direct Method, Short Shortcut method, & Step deviation method.
- Direct method:If x1, x2, ... , xn has corresponding f1, f2, … fn frequencies, then the arithmetic mean is given by, X= (\(\sum_{i=1}^{n}\)fi xi) / (\(\sum_{i=1}^{n}\)nfi).
- Shortcut Method: X = A + (1/N)(\(\sum_{i=1}^{n}\)fi di), where A is an arbitrary point and di = xi - A.
- Step deviation method: X = A +h(1/N) \(\sum_{i=1}^{n}\)fi ui , ui = (xi - A)/h
Sample Questions
Ques. The heights of five students are 5.0 ft, 6.0 ft, 4.6ft, 5.5ft, and 6.2ft respectively. Using the arithmetic mean formula, find the average height of the students. (2 Marks)
Solution:
Average height of the students is,
Arithmetic mean = {Sum of Observation}/{Total numbers of Observations}
Or, (5.0 + 6.0 + 4.6 + 5.5 + 6.2)/5
Or, 27.3/5 = 5.46ft.
Therefore, the average height of the students is 5.46ft.
Ques. The arithmetic mean of (2x + 1), (x + 2), (3x), (4x + 4) is x + 1, find the value of x. (3 Marks)
Solution:
X = [(2x + 1) + (x + 2) + (3x) + (4x + 4)] / 4
X = (10x + 7) / 4
But it’s given that arithmetic mean is x + 1
Or, x + 1 = (10x + 7) / 4
Or, 4 (x + 1) = (10x + 7)
Or, 4x + 4 = 10x + 7
Or, −6x = 3
Or, x = -1/2
Therefore the value of x is -1/2
Ques. Find the arithmetic mean of the following distribution: (5 Marks)
x: 82 50 55 10 90
f: 2 1 3 2 1
Solution:
| xi | fi | fi xi |
| 82 50 55 10 90 | 2 1 3 | 2 1 | 164 50 165 20 90 |
| fi = 9 | fi xi =489 |
Using the formula,
X= (i=1nfi xi) / (i=1nfi)
X = 489 / 9 = 54.3
Therefore, the mean of the distribution is 54.3
Ques. Find the value of x if the arithmetic mean of the following distribution is 6. (5 Marks)
x: 2 4 6 10 x+5
f: 5 2 10 7 8
Solution:
| xi | fi | fi xi |
| 2 4 6 1 x+5 | 5 2 10 7 8 | 10 8 60 7 8x+40 |
| \(\sum\)fi = 32 | \(\sum\)fi xi =8x+125 |
Using the formula,
X= (\(\sum_{i=1}^{n}\)fi xi) / (\(\sum_{i=1}^{n}\)fi)
X = (8x +125) / 32
But X = 6 then,
Or, 6 (32) = 8x + 125
Or, 256 = 8x + 125
Or, x= 131/8 = 16.3
Ques. Find the value of x if the arithmetic mean of the following distribution is 2. (5 Marks)
x: 3 5 7 19 11 13
f: 6 8 15 x 8 4
Solution:
| xi | fi | fi xi |
| 3 5 7 19 11 13 | 6 8 15 x 8 4 | 24 40 105 9x 88 54 |
| \(\sum\)fi = 41 + x | \(\sum\)fi xi = 303+9x |
Using the formula,
X= (\(\sum_{i=1}^{n}\)fi xi) / (\(\sum_{i=1}^{n}\)fi)
X = (9x + 303) / (41 + x)
But X = 8 then,
Or, 8 (41 + x) = 9x + 303
Or, 328 +8x = 9x + 303
Or, x= 25
Ques. Find the value of x if the arithmetic mean of the following distribution is 2. (5 Marks)
x: 1 x 5 9 10
f: 5 2 1 6 5
Solution:
| xi | fi | fi xi |
| 1 x 5 9 10 | 5 1 1 6 5 | 5 x 5 54 60 |
| fi = 19 | fi xi =x+124 |
Using the formula,
X= (\(\sum_{i=1}^{n}\)fi xi) / (\(\sum_{i=1}^{n}\)fi)
X = (2x +124) / 19
But X = 7 then,
Or, 133 = x + 124
Or, 9 = x + 125
Or, x= 9
Ques. If the arithmetic mean of the following distribution is 1.4 and total frequency is 200, find the values of f1 and f2. (5 Marks)
x: 0 1 2 3 4 5
f: 46 f1 f2 25 10 5
Solution:
| xi | fi | fi xi |
| 0 | 46 f1 f2 25 10 5 | 0 f1 2*f2 75 40 25 |
| \(\sum\)fi = 86 + f1 + f2 | \(\sum\)fi xi =f1 + 2f2 +140 |
\(\sum\)fi = 86 + f1 + f2 = N =200
f1 + f2 = 114 ….(i)
Using the formula,
X= (\(\sum_{i=1}^{n}\) fi xi) / (\(\sum_{i=1}^{n}\)fi)
But X = 1.4 then,
Or, 1.4(200) = f1 + 2f2 +140
Or, f1 + 2f2 = 152 …..(ii)
Solving (i) and (ii)
f1 = 76
f2 = 38
Ques. Find the Arithmetic mean of the following distribution table: (5 Marks)
xi: 800 820 860 900 920 980 1000
fi: 7 14 19 25 20 10 5
Solution:
Let assume A = 900
| xi | fi | di = xi - A or xi - 900 | fi di |
| 800 820 860 900 920 980 1000 | 7 14 19 25 20 10 5 | -100 -80 -40 0 20 80 100 | -700 -1120 -760 0 400 800 500 |
| N= \(\sum\)fi = 100 | \(\sum\)fi di = -880 |
Using the formula, we get
X = A + (1/N) (\(\sum_{i=1}^{n}\) fi di)
X = 900 + (1/100)(-880) = 891.2
Therefore the mean is 891.2.
Ques. Find the Arithmetic mean of the following distribution table: (5 Marks)
Weight: 60 77 72 73 75
No. of students: 4 3 2 2 1
Solution:
Let assume A = 72
| xi | fi | di = xi - A xi - 72 | fi di |
| 60 77 72 73 75 | 4 3 3 2 1 | -12 | -48 15 0 2 3 |
| N= \(\sum\)fi = 12 | \(\sum\)fi di = -28 |
Using the formula, we get
X = A + (1/N) (\(\sum_{i=1}^{n}\) fi di)
X = 72 + (1/12)(-28) = 69.6kg
Ques. Find the average of the following distribution table: (5 Marks)
Marks obtained: 81 84 86 90 98 99 100
No. of students: 7 1 1 2 2 1 0
Solution:
Let A = 90 and h = 2
| xi | fi | di = xi - A = xi - 90 | ui = (xi - 90)/2 | fi ui |
| 81 84 86 90 98 99 100 | 7 1 1 2 2 1 0 | -9 -6 -4 0 8 9 10 | -4.5 -3 -2 0 4 4.5 5 | -31.5 -3 -2 0 8 9 0 |
| \(\sum\)fi = 14 | \(\sum\)fi ui =-19.5 |
Using the formula, we get,
X = A +h(1/N) \(\sum_{i=1}^{n}\) fiui
X = 90 + (2/14) (-19.5) = 900 -8.80 =87.2
Therefore, the average of the given data distribution is 87.2.
Ques. Find the average of the following distribution table using the step deviation method: (3 Marks)
Height: 200 600 1000 1400 1800 2200
No. of village: 142 265 560 251 271 89
Solution:
Let A = 1400 and h = 400
| xi | fi | di = xi - A = xi - 1400 | ui = (xi - 1400)/400 | fi ui |
| 200 600 1000 1400 1800 2200 | 142 265 560 251 271 89 | -1200 -800 -400 0 400 800 | -3 -2 -1 0 1 2 | -426 -530 -560 0 89 32 |
| \(\sum\)fi = 1343 | \(\sum\)fi ui=-1395 |
Using the formula, we get,
X = A +h(1/N) \(\sum_{i=1}^{n}\) fiui
X = 1400 + (400/1343) (-1395) = 1400 -415.4 = 984.6
Ques. Find the average of the following distribution table using the Step deviation method. (3 Marks)
Wage: 800 820 860 900 920 980 1000
No. of workers: 7 14 19 25 20 10 5
Solution:
Let A = 900 and h = 20
| xi | fi | di = xi - A = xi - 900 | ui = (xi - 900)/20 | fi ui |
| 800 820 860 900 920 980 1000 | 7 14 19 25 20 10 5 | -100 100 | -5 | -35 -56 -38 0 20 40 25 |
| \(\sum\)fi = 100 | \(\sum\)fi ui = -44 |
Using the formula, we get,
X = A +h(1/N) \(\sum_{i=1}^{n}\) fiui
X = 900 + (20/100) (-44) = 900 -8.80 =891.2
Therefore, the average of the given data distribution is 891.2.
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