**Arithmetic Mean**

On the basis of the type of data series that has provided to us (ie, Individual, Discrete, Continuous), it will be convenient if we use appropriate formula for finding averages in each of these series.

There are three methods by which Simple mean can be calculated in each of these three series.

They are :

**Direct Method****Assumed Mean Method****Step Deviation Method**

**Calculation of Arithmetic Mean**

Calculation of arithmatic mean can be studied under two heads.

**Arithmetic Mean for Ungrouped Data****Arithmetic Mean for Grouped Data**

**Arithmetic Mean for Ungrouped Data**

Arithmetic Mean for Ungrouped data can be calculated using the following methods:

**Direct Method****Assumed Mean Method****Step Deviation Method**

**Individual Series**

**Direct Method**

STEPS

- Find the sum of observations (∑X)
- Take the number of observations (N)
- Use the formula \( X̅ = {{{\frac{ΣX}{N}} }} \)

$$ A.M = {{{\frac{60+75+72+68+80+65}{6}} }} $$ $$ = {{{\frac{420}{6}} }} = 70 $$

**Assumed Mean Method**

STEPS

- Take an assumed mean A
- Take the deviation of each X from the assumed mean. ie., d = X - A
- Find the sum of the deviations to get Σd
- Use the formula \( X̅ = A + {{{\frac{Σd}{N}} }} \) ; where N is the total number of observations

2500, 6500, 3000, 5500, 4500, 6000, 3500, 3000, 5500, 5000, 2000, 4500, 3500, 3000, 4500, 6500, 4000, 3000, 2500, 4500

Table 5.1 | |
---|---|

X | d=X-A=X-4000 |

2500 | -1500 |

6500 | 2500 |

3000 | -1000 |

5500 | 1500 |

4500 | 500 |

6000 | 2000 |

3500 | -500 |

3000 | -1000 |

5500 | 1500 |

5000 | 1000 |

2000 | -2000 |

4500 | 500 |

3500 | -500 |

3000 | -1000 |

4500 | 500 |

6500 | 2500 |

4000 | 0 |

3000 | -1000 |

2500 | -1500 |

4500 | 500 |

N=20 | Σd = (-10000)+13000=3000 |

$$=4000 + 150 = 4150 $$

**Step Deviation Method**

Complexity of calculations in finding arithmatic mean can nfurther be reduced by using step deviation method.

STEPS

- Take an assumed mean A
- Take the deviation of each X from the assumed mean. ie., d = X - A
- Devide the deviation d by common factor c, i.e., \( d ' = {{{\frac{d}{c}} }} = {{{\frac{X - A}{c}} }} \)
- Then find Σd’
- Use the formula \( X̅ = A + {{{\frac{Σd'}{N}} }} × c \) ; where N is the total number of observations

45, 30, 65, 70, 40, 25, 45, 25, 55, 40, 20, 50

Assumed mean is taken as 50.

Let c = 5 (we decide value of c only after seeing the column for d in the table)

Table 5.2 | ||
---|---|---|

X | d=X-A=X-50 | \( \mathbf {d ' = {{{\frac{d}{c}} }} = {{{\frac{d}{5}} }}} \) |

45 | -5 | -1 |

30 | -20 | -4 |

65 | 15 | 3 |

70 | 20 | 4 |

40 | -10 | -2 |

25 | -25 | -5 |

45 | -5 | -1 |

25 | -25 | -5 |

55 | 5 | 1 |

40 | -10 | -2 |

20 | -30 | -6 |

50 | 0 | 0 |

N = 12 | Σd’ = (-26) + 8 =-18 |

$$ X̅ = A + {{{\frac{Σd'}{N}} }} × c = {50+\frac{-18}{12}} × 5 $$

$$= 42.5 $$

**Arithmetic Mean for Grouped Data**

**Discrete Series**

**Direct Method**

STEPS

- Multiply the frequency against each observation with the value of observation to get fx
- Add the column of fx to get Σfx
- Use the formula \( X̅ = {{{\frac{Σfx}{Σf}} }} \)

Table 5.3 | |
---|---|

Income | Number of Persons |

1200 | 2 |

1500 | 10 |

1800 | 15 |

2000 | 7 |

2300 | 5 |

2600 | 4 |

3400 | 3 |

4200 | 3 |

5000 | 1 |

Now we can create a table to find fx.

Table 5.4 | ||
---|---|---|

Income | Number of Persons | fX |

1200 | 2 | 2400 |

1500 | 10 | 15000 |

1800 | 15 | 27000 |

2000 | 7 | 14000 |

2300 | 5 | 11500 |

2600 | 4 | 10400 |

3400 | 3 | 10200 |

4200 | 3 | 12600 |

5000 | 1 | 5000 |

Σf = 50 | Σfx = 108100 |

$$ X̅ = {{{\frac{Σfx}{Σf}} }} = {\frac{-108100}{50}} $$

$$= 2162 $$

**Assumed Mean Method**

Assumed mean method is applying just to simplify the calculations.

STEPS

- Take an assumed mean A
- Take the deviation ‘d’ of each X from the assumed mean. ie., d = X - A
- Multiply d with f to get fd
- Add the column of fd to get Σfd
- Add all the frequencies to get Σf
- Use the formula \( X̅ = A + {{{\frac{Σfd}{Σf}} }} \)

Table 5.5 | |
---|---|

Income | Number of Persons |

1200 | 2 |

1500 | 10 |

1800 | 15 |

2000 | 7 |

2300 | 5 |

2600 | 4 |

3400 | 3 |

4200 | 3 |

5000 | 1 |

Now we can create a table to find d and fd.

Table 5.6 | |||
---|---|---|---|

Income | Number of Persons | d = X-A = X-2300 | fd |

1200 | 2 | -1100 | -2200 |

1500 | 10 | -800 | -8000 |

1800 | 15 | -500 | -7500 |

2000 | 7 | -300 | -2100 |

2300 | 5 | 0 | 0 |

2600 | 4 | 300 | 1200 |

3400 | 3 | 1100 | 3300 |

4200 | 3 | 1900 | 5700 |

5000 | 1 | 2700 | 2700 |

Σf = 50 | Σfd=(-19800)+12900 = -6900 |

$$ X̅ = A + {{{\frac{Σfd}{Σf}} }} = 2300 + {\frac{-6900}{50}} $$

$$= 2162 $$

**Step Deviation Method**

Complexity of calculations in finding arithmatic mean can nfurther be reduced by using step deviation method. Under step deviation method, as in the individual series case, we divide the deviations d by a common factor say c. We denote that as d’.

STEPS

- Take an assumed mean A
- Take the deviation of each X from the assumed mean. ie., d = X - A
- Devide the deviation d by common factor c, i.e., \( d ' = {{{\frac{d}{c}} }} = {{{\frac{X - A}{c}} }} \)
- Multiply each d’ with frequency to get fd’
- Add the column of fd’ to get Σfd’
- Add all the frequencies to get Σf
- Use the formula \( X̅ = A + {{{\frac{Σfd'}{Σf}} }} × c \) ; where N is the total number of observations

Table 5.7 | |
---|---|

Income | Number of Persons |

1200 | 2 |

1500 | 10 |

1800 | 15 |

2000 | 7 |

2300 | 5 |

2600 | 4 |

3400 | 3 |

4200 | 3 |

5000 | 1 |

Now we can create a table to find d, d' and fd'. We can take assumed mean as 2300 and c as 100. C is taken as 100 after getting d column.

Table 5.8 | ||||
---|---|---|---|---|

X | f | d = X-A = X-2300 | \( \mathbf {d ' = {{{\frac{d}{c}} }} = {{{\frac{d}{100}} }}} \) | fd’ |

1200 | 2 | -1100 | -11 | -22 |

1500 | 10 | -800 | -8 | -80 |

1800 | 15 | -500 | -5 | -75 |

2000 | 7 | -300 | -3 | -21 |

2300 | 5 | 0 | 0 | 0 |

2600 | 4 | 300 | 3 | 12 |

3400 | 3 | 1100 | 11 | 33 |

4200 | 3 | 1900 | 19 | 57 |

5000 | 1 | 2700 | 27 | 27 |

Σf = 50 | Σfd’ = (-198) + 129 = -69 |

$$= 2162 $$

**Continuous Series**

STEPS

- Find the mid value of each class, denoted by m
- Multiply each class frequency with m to get fm
- Add the column fm to get Σfm
- Use the formula \( X̅ = {{{\frac{Σfm}{Σf}} }} \)

# Note :-

- For finding arithmetic mean, it is not necessary to convert inclusive type to exclusive type. However, in case of finding median or; mode the coiwersion .is necessary-
- If intervals are having unequal width, then also the same method can be followed.
- If the distribution is exdlusive type, then, the class interval ot a class is got by the formula (upper limit - lower limit)
- If the distribution is inclusive type, then, there is a gap between the upper limit of the class and the lower limit of the next class. In that case, the class interval of the class is got by the formula, (upper limit - lower limit + gap).

**Direct Method**

Table 5.9 | |
---|---|

Marks | Number of Students |

0-10 | 8 |

10-20 | 12 |

20-30 | 15 |

30-40 | 17 |

40-50 | 25 |

50-60 | 20 |

60-70 | 16 |

70-80 | 13 |

80-90 | 10 |

90-100 | 4 |

Let us create a table showing m and fm.

Table 5.10 | |||
---|---|---|---|

Marks | Number of Students | m | fm |

0-10 | 8 | 5 | 40 |

10-20 | 12 | 15 | 180 |

20-30 | 15 | 25 | 375 |

30-40 | 17 | 35 | 595 |

40-50 | 25 | 45 | 1125 |

50-60 | 20 | 55 | 1100 |

60-70 | 16 | 65 | 1040 |

70-80 | 13 | 75 | 975 |

80-90 | 10 | 85 | 850 |

90-100 | 4 | 95 | 380 |

Σf = 140 | Σfm = 6660 |

$$ X̅ = {{{\frac{Σfm}{Σf}} }} = {\frac{6660}{140}} $$

$$= 47.57 $$

**Assumed Mean Method**

Assumed mean method is applying just to simplify the calculations.

STEPS

- Find the mid value (m) of each class
- Take an assumed mean, A
- Take deviation of each m from A, i.e., d = m - A
- Multiply d with f to get fd
- Add all fd to get Σfd
- Add all f to get Σf
- Apply the formula \( X̅ = A + {{{\frac{Σfd}{Σf}} }} \)

Table 5.11 | |
---|---|

Marks | Number of Students |

0-10 | 8 |

10-20 | 12 |

20-30 | 15 |

30-40 | 17 |

40-50 | 25 |

50-60 | 20 |

60-70 | 16 |

70-80 | 13 |

80-90 | 10 |

90-100 | 4 |

Let us create a table showing m, d and fd. Assumed mean is taken as 45.

Table 5.12 | ||||
---|---|---|---|---|

Marks | Number of Students | m | d = m-45 | fd |

0-10 | 8 | 5 | -40 | -320 |

10-20 | 12 | 15 | -30 | -360 |

20-30 | 15 | 25 | -20 | -300 |

30-40 | 17 | 35 | -10 | -170 |

40-50 | 25 | 45 | 0 | 0 |

50-60 | 20 | 55 | 10 | 200 |

60-70 | 16 | 65 | 20 | 320 |

70-80 | 13 | 75 | 30 | 390 |

80-90 | 10 | 85 | 40 | 400 |

90-100 | 4 | 95 | 50 | 200 |

Σf = 140 | Σfd = -1150 + 1510 = 360 |

$$ X̅ = A + {{{\frac{Σfd}{Σf}} }} = 45 + {\frac{360}{140}} $$

$$= 47.57 $$

**Step Deviation Method**

In this method, in order to simplify calculations we take a common factor for the data.

STEPS

- Find the mid value (m) of each class
- Take an assumed mean A
- Take the deviation of each m from A, i.e., d = m-A
- Divide d by a common factor c to get d’. \( d ' = {{{\frac{d}{c}} }} = {{{\frac{m - A}{c}} }} \)
- Multiply d’ with f to get Σfd’
- Add all the frequencies to get Σf
- Apply the formula \( X̅ = A + {{{\frac{Σfd'}{Σf}} }} × c \)

**EXCLUSIVE CLASS**

Table 5.13 | ||||
---|---|---|---|---|

Marks | Number of Students | |||

0-10 | 8 | |||

10-20 | 12 | |||

20-30 | 15 | |||

30-40 | 17 | |||

40-50 | 25 | |||

50-60 | 20 | |||

60-70 | 16 | |||

70-80 | 13 | |||

80-90 | 10 | |||

90-100 | 4 |

To find arithmetic mean through step deviation method, we need to find m, d, d' and fd'.

Table 5.14 | |||||
---|---|---|---|---|---|

X | f | m | d = m-45 | \( \mathbf {d ' = {{{\frac{d}{c}} }} = {{{\frac{d}{10}} }}} \) | fd' |

0-10 | 8 | 5 | -40 | -4 | -32 |

10-20 | 12 | 15 | -30 | -3 | -36 |

20-30 | 15 | 25 | -20 | -2 | -30 |

30-40 | 17 | 35 | -10 | -1 | -17 |

40-50 | 25 | 45 | 0 | 0 | 0 |

50-60 | 20 | 55 | 10 | 1 | 20 |

60-70 | 16 | 65 | 20 | 2 | 32 |

70-80 | 13 | 75 | 30 | 3 | 39 |

80-90 | 10 | 85 | 40 | 4 | 40 |

90-100 | 4 | 95 | 50 | 5 | 20 |

Σf = 140 | Σfd’ = (-115) + 151 = 36 |

$$ X̅ = A + {{{\frac{Σfd'}{Σf}} }} × c = 45 + {\frac{36}{140}} × 10 $$

$$= 47.57 $$

**INCLUSIVE CLASS**

Find arithmetic mean of the following distribution using step deviation method. The given distribution is in inclusive format.

Table 5.15 | |||||
---|---|---|---|---|---|

Marks | Number of Students | ||||

0-9 | 1 | ||||

10-19 | 5 | ||||

20-29 | 12 | ||||

30-39 | 15 | ||||

40-49 | 24 | ||||

50-59 | 20 | ||||

60-69 | 9 | ||||

70-79 | 8 | ||||

80-89 | 4 | ||||

90-99 | 2 |

Here the classes are inclusive type. It is not neccessary to convert them into exclusive, because mid-points remain the same whether or not the adjustment is made. To find arithmetic mean through step deviation method, we need to find m, d, d' and fd'

Table 5.16 | |||||
---|---|---|---|---|---|

X | f | m | d = m-45 | \( \mathbf {d ' = {{{\frac{d}{c}} }} = {{{\frac{d}{10}} }}} \) | fd' |

0-9 | 8 | 4.5 | -50 | -5 | -5 |

10-19 | 12 | 14.5 | -40 | -4 | -20 |

20-29 | 15 | 24.5 | -30 | -3 | -36 |

30-39 | 17 | 34.5 | -20 | -2 | -30 |

40-49 | 25 | 44.5 | -10 | -1 | -24 |

50-59 | 20 | 54.5 | 0 | 0 | 0 |

60-69 | 16 | 64.5 | 10 | 1 | 9 |

70-79 | 13 | 74.5 | 20 | 2 | 16 |

80-89 | 10 | 84.5 | 30 | 3 | 12 |

90-99 | 4 | 94.5 | 40 | 4 | 8 |

Σf = 140 | Σfd’ = (-115) + 45 = -70 |

$$ X̅ = A + {{{\frac{Σfd'}{Σf}} }} × c = 54.5 + {\frac{-70}{100}} × 10 $$

$$= 47.57 $$

**UNEQUAL CLASSES**

Table 5.17 | |||||
---|---|---|---|---|---|

Marks | Number of Students | ||||

0-5 | 8 | ||||

5-15 | 12 | ||||

15-20 | 15 | ||||

20-40 | 17 | ||||

40-50 | 25 | ||||

50-60 | 20 | ||||

60-65 | 16 | ||||

65-80 | 13 | ||||

80-90 | 10 | ||||

90-100 | 4 |

To find arithmetic mean through step deviation method, we need to find m, d, d' and fd'.

Table 5.18 | |||||
---|---|---|---|---|---|

X | f | m | d = m-45 | \( \mathbf {d ' = {{{\frac{d}{c}} }} = {{{\frac{d}{2.5}} }}} \) | fd' |

0-5 | 8 | 2.5 | -52.5 | -21 | -168 |

5-15 | 12 | 10 | -45 | -18 | -216 |

15-20 | 15 | 17.5 | -37.5 | -15 | -225 |

20-40 | 17 | 30 | -25 | -10 | -170 |

40-50 | 25 | 45 | -10 | -4 | -100 |

50-60 | 20 | 55 | 0 | 0 | 0 |

60-65 | 16 | 62.5 | 7.5 | 3 | 48 |

65-80 | 13 | 72.5 | 17.5 | 7 | 91 |

80-90 | 10 | 85 | 30 | 12 | 120 |

90-100 | 4 | 95 | 40 | 16 | 64 |

Σf = 140 | Σfd’ = (-879) + 323 = -556 |

$$= 45.07 $$

**OPEN END CLASSES**

Table 5.19 | |||||
---|---|---|---|---|---|

Marks | Number of Students | ||||

Less than 15 | 8 | ||||

15-30 | 15 | ||||

30-45 | 32 | ||||

45-60 | 26 | ||||

60-75 | 13 | ||||

Above 75 | 6 |

Here the given table has two open end classes, namely. the first and the last. In the case of open end classes, we cannot find out the arithmetic mean unless we make an assumption about the limits of the open end classes. For that, look at the class intervals of the classes following the first class and preceding the last class. The class interval of the second class is 15. Hence by assuming the class interval of the first class also as 15, we get the lower limit of the first class as 0. Similarly, since the class interval of the class preceding to the last class is 15, by assuming the class interval of the last class as 15, we get the upper limit of the last class as 90.

Now we get the adjusted table as given below.

Table 5.20 | |||||
---|---|---|---|---|---|

Marks | Number of Students | ||||

0-15 | 8 | ||||

15-30 | 15 | ||||

30-45 | 32 | ||||

45-60 | 26 | ||||

60-75 | 13 | ||||

75-90 | 6 | ||||

Total | 100 |

Now we can simply find arithmetic mean by using any method. Here , we used step deviation method to find arithmetic mean.

Table 5.21 | |||||
---|---|---|---|---|---|

X | f | m | d = m-52.5 | \( \mathbf {d ' = {{{\frac{d}{c}} }} = {{{\frac{d}{5}} }}} \) | fd' |

0-15 | 8 | 7.5 | -45 | -9 | -72 |

15-30 | 12 | 22.5 | -30 | -6 | -90 |

30-45 | 15 | 37.5 | -15 | -3 | -96 |

45-60 | 17 | 52.5 | 0 | 0 | 0 |

60-75 | 25 | 67.5 | 15 | 3 | 39 |

75-90 | 20 | 82.5 | 30 | 6 | 36 |

Σf = 100 | Σfd’=(-258)+75 = -183 |

$$= 43.35 $$

**MORE THAN CUMULATIVE FREQUENCIES**

Table 5.22 | |||||
---|---|---|---|---|---|

Wage | Number of Labourers | ||||

Above 0 | 675 | ||||

Above 10 | 625 | ||||

Above 20 | 550 | ||||

Above 30 | 450 | ||||

Above 40 | 275 | ||||

Above 50 | 150 | ||||

Above 60 | 75 | ||||

Above 70 | 25 |

Here the frequencies given are not the simple frequencies. They are more than cumulative frequencies. Hence we need to convert them to simple frequency. The simple frequency of the first class = 675 — 625 = 50. For the second class, simple frequency = 625 — 550 = 75. For the third class, simple frequency = 550 — 450 = 100; and so on. Also we should write the classes accordingly.

Now we get the adjusted table as given below.

Table 5.23 | |||||
---|---|---|---|---|---|

X | f | ||||

0-10 | 50 | ||||

10-20 | 75 | ||||

20-30 | 100 | ||||

30-40 | 175 | ||||

40-50 | 125 | ||||

50-60 | 75 | ||||

60-70 | 50 | ||||

70-80 | 25 | ||||

Total | 675 |

Now we can find arithmetic mean using step deviation method.

Table 5.24 | |||||
---|---|---|---|---|---|

X | f | m | d = m-35 | \( \mathbf {d ' = {{{\frac{d}{c}} }} = {{{\frac{d}{10}} }}} \) | fd' |

0-10 | 50 | 5 | -30 | -3 | -150 |

10-20 | 75 | 15 | -20 | -2 | -150 |

20-30 | 100 | 25 | -10 | -1 | -100 |

30-40 | 175 | 35 | 0 | 0 | 0 |

40-50 | 125 | 45 | -10 | 1 | 125 |

50-60 | 75 | 55 | 20 | 2 | 150 |

60-70 | 50 | 65 | 30 | 3 | 150 |

70-80 | 25 | 75 | 40 | 4 | 100 |

Σf = 675 | Σfd’=(-400)+525 = 125 |

$$= 36.85 $$

**LESS THAN CUMULATIVE FREQUENCIES**

Table 5.25 | |||||
---|---|---|---|---|---|

Mark | Number of Students | ||||

Less than 10 | 4 | ||||

Less than 20 | 16 | ||||

Less than 30 | 40 | ||||

Less than 40 | 76 | ||||

Less than 50 | 96 | ||||

Less than 60 | 112 | ||||

Less than 70 | 120 | ||||

Less than 80 | 125 |

By inspection, we can see that the frequencies given are not the simple frequencies. They are less than cumulative frequencies. In order to calculate arithmetic mean, we need simple frequency of each class. So first we need to convert cumulative frequencies to simple frequency. The simple frequency of the first class is 4 itself. For the second class, simple frequency = 16 — 4 = 12. For the third class, simple frequency = 40 — 16 = 24; and so on. Also we should write the classes accordingly.

Now we get the adjusted frequencies as in the table given below.

Table 5.26 | |||||
---|---|---|---|---|---|

X | f | ||||

0-10 | 4 | ||||

10-20 | 12 | ||||

20-30 | 24 | ||||

30-40 | 36 | ||||

40-50 | 20 | ||||

50-60 | 16 | ||||

60-70 | 8 | ||||

70-80 | 5 | ||||

Total | 125 |

Now we can find arithmetic mean using step deviation method.

Table 5.27 | |||||
---|---|---|---|---|---|

X | f | m | d = m-35 | \( \mathbf {d ' = {{{\frac{d}{c}} }} = {{{\frac{d}{10}} }}} \) | fd' |

0-10 | 4 | 5 | -30 | -3 | -12 |

10-20 | 12 | 15 | -20 | -2 | -24 |

20-30 | 24 | 25 | -10 | -1 | -24 |

30-40 | 36 | 35 | 0 | 0 | 0 |

40-50 | 20 | 45 | 10 | 1 | 20 |

50-60 | 16 | 55 | 20 | 2 | 32 |

60-70 | 8 | 65 | 30 | 3 | 24 |

70-80 | 5 | 75 | 40 | 4 | 20 |

Σf = 125 | Σfd’=(-60)+96 = 36 |

$$ X̅ = A + {{{\frac{Σfd'}{Σf}} }} × c = 35 + {\frac{36}{125}} × 10 $$

$$= 37.88 $$

**COMBINED MEAN**

Consider a sample 20, 23, 15, 21, 28 and 19. We can call this as sample - I. Its arithmetic mean is :

#### \( {\frac{20+23+15+21+28+19}{6}} \)

\( = {\frac{126}{6}} = 21 \)

#### \( {\frac{32+28+19+13}{4}} \)

\( = {\frac{92}{4}} = 23 \)

#### \( {\frac{20+23+15+21+28+19+32+28+19+13}{10}} \)

\( = {\frac{218}{10}} \)

\( = 21.8 \)

_{1}and size of sample-II as n

_{2}, ; Also, we denote the mean of sample- I as X̄

_{1}and mean of sample- II as X̄

_{2},. Then the combined mean X̄ of samples.1 and 2 can be determined using the formula:

#### \( X̅ = {\frac{n_1 X̅_1 + n_2 X̅_2}{n_1 + n_2}} \)

_{1}= 6, n

_{2}= 4, X̅

_{1}= 21 and X̅

_{2}= 23.

#### \( X̅ = {\frac{n_1 X̅_1 + n_2 X̅_2}{n_1 + n_2}} \)

\( = {\frac{6 × 21 + 4 × 23}{6 + 4}} \)

\( = {\frac{126 + 92}{10}}\)

\( = {\frac{126 + 92}{10}}\)

\( = 21.8\)

**CORRECTION IN MEAN**

While calculating mean, sometimes we may consider numbers wrongly by mistake. This will lead to wrong results. But, when we realized the mistake, the results may be corrected without doing the problem afresh. The process of finding out the correct mean value is very simple. First we multiply the incorrect mean by the total number of observations. The incorrect values are then subtracted from that and the correct values are added, Then divide the number you have got by the total number of observations. The below given example will illustrate the method.

The average mark secured by 50 students was calculated as 48. later on, it was found that a mark 60 was misred as 16. Find the correct average mark secured by the students.

#### Incorrect mean = 48

#### Incorrect mean × Number of observations

#### = 48 × 150 = 7200

#### Subtracting incorrect value, 7184 - 16 = 7184

#### Adding correct value, 7184 + 60 = 7244

$$ { {{Correct} \, {mean =}}\frac{7244}{150} = 48.3} $$

**WEIGHTED MEAN**

The arithmetic mean we had studied is simple arithmetic mean. In the calculation of simple arithmetic mean each item of the series is considered equally important. But there may be cases where all items may not have equal importance. Some of them may be comparatively more important than others. For example, if we are finding out the change in the cost of living of a certain group of people and if we merely find the simple arithmetic average of the prices of the commodities consumed by them, the average would be unrepresentative. All the items of consumption are not equally important. The price of salt may increase by 100 per cent but this will not affect the cost of living to the extent to which it would be affected, if the price of rice goes up only by 10 per cent. In such cases simple average is not suitable and we give different weights to each item while computing average. Arithmetic mean computed by assigning different weights to each item is called weighted arithmetic mean.

Let x_{1}, x_{2}, .... , x_{n} be n items with weights w_{1}, w_{2}, ... , w_{n} respectively. Then the weighted arithmetic mean is:

$$ { X̅_w =\frac{w_1 x_1 + w_2 x_2 +...+ w_n x_n}{w_1 + w_2 +...+ w_n} = \frac{Σwx}{Σw}} $$

Let us find weighted arithmatic mean of the given data.

Table 5.28 | |||||
---|---|---|---|---|---|

Item | Amount | Weight | |||

1 | 35 | 3 | |||

2 | 27 | 7 | |||

3 | 65 | 1 | |||

4 | 47 | 4 | |||

5 | 30 | 9 |

Now we need to create a table with w and wx as given below.

Table 5.29 | |||||
---|---|---|---|---|---|

x (Amount) | w (Weight) | wx | |||

35 | 3 | 105 | |||

27 | 7 | 189 | |||

65 | 1 | 65 | |||

47 | 4 | 188 | |||

30 | 9 | 270 | |||

Σw 24 = 24 | Σwx = 817 |

$$ { X̅_w =\frac{Σwx}{Σw} = \frac{817}{24}} = 34.04 $$

Let us practice an another example.

An examination in the subjects, English, Mathematics and Social Science was held to decide the award of a scholarship. The marks obtained by the top three candidates are given below. We can find out who is going to win the scholarship.We can find out the possibility of change in the selection of student, if we use simple arithmetic mean for selection.

Table 5.30 | |||||
---|---|---|---|---|---|

Subject | Student 1 | Student 2 | Student 3 | Weight | |

English | 49 | 44 | 45 | 3 | |

Mathematics | 42 | 42 | 50 | 5 | |

Social Science | 48 | 50 | 45 | 2 |

Table 5.31 | ||||||||
---|---|---|---|---|---|---|---|---|

Student 1 | Student 2 | Student 3 | ||||||

X | w | wx | X | w | wx | X | w | wx |

49 | 3 | 147 | 44 | 3 | 132 | 42 | 3 | 126 |

42 | 5 | 210 | 42 | 5 | 210 | 50 | 5 | 250 |

48 | 2 | 96 | 50 | 2 | 100 | 45 | 2 | 90 |

139 | Σw=10 | Σwx=453 | 136 | Σw=10 | Σwx=442 | 137 | Σw=10 | cwx=442 |

Table 5.32 | ||||||||
---|---|---|---|---|---|---|---|---|

Student 1 | Student 2 | Student 3 | ||||||

Weighted mean $$ { X̅_w =\frac{Σwx}{Σw}} $$ $$ {= \frac{453}{10}} = 45.3 $$ |
Weighted mean $$ { X̅_w =\frac{Σwx}{Σw} } $$ $$ { = \frac{442}{10}} = 44.2 $$ |
Weighted mean $$ { X̅_w =\frac{Σwx}{Σw}} $$ $$ { = \frac{466}{10}} = 46.6 $$ |
||||||

Simple mean $$ { X̅ =\frac{Σx}{N} } $$ $$ {= \frac{139}{3}} = 46.3 $$ |
Simple mean $$ { X̅ =\frac{Σx}{N} } $$ $$ {= \frac{136}{3}} = 45.3 $$ |
Simple mean $$ { X̅ =\frac{Σx}{N} } $$ $$ { = \frac{137}{3}} = 45.7 $$ |

**AN INTERESTING PROPERTY OF AM**

It is interesting to note that the sum of deviations of items in a series about arithmetic mean is always equal to zero.

Symbolically, Σ(X - X̅) = 0

However, arithmetic mean is affected by extreme items. That is, any large value on either end, can push it up or down.

Consider 6 items 20, 23, 15, 21, 28 and 19. Its arithmetic mean is 21. Their deviation from the arithmetic mean are -1, 2, -6, 0, 7, and -2 respectively.

Then, the sum of the deviations = (-1) + 2 + (-6) + 0 + 7 + (-2) = 0

**Merits of Arithmetic Mean**

- It is the most popular average
- It is easy-to understand
- It is based on all items of the series
- It is not very much affected by sampling fluctuations
- It is capable of further algebraic treatment
- It is not a positional value
- It is rigidly defined so as to avoid ambiguity

**Demerits of Arithmetic Mean**

- It cannot be determined by inspection
- It cannot be determined graphically
- It is affected by extreme values
- It is not suitable for averaging ratios or percentages
- It cannot be determined if one of the observations is missing
- It may not be a number in the series
- For open-end classes, assumptions may be made about the class interval