In statistics, assumed mean method is used for calculating mean or arithmetic mean of a grouped data. If the given data is large, then this method is recommended rather than a direct method for calculating mean. This method helps in reducing the calculations and results in small numerical values for easy computation. This method depends on estimating the mean and rounding to an easy value to calculate with. Again this value is subtracted from all the sample values. When the samples are converted into equal size ranges or class intervals, a central class is chosen and the count of ranges from that is used in the calculations.

## Assumed Mean Method Formula

Let x_{1}, x_{2}, x_{3},…,x_{n} are mid-ponts or class marks of n class intervals and f_{1}, f_{2}, f_{3}, â€¦, f_{n} are the respective frequencies. The formula of assumed mean method is:

Here,

a = assumed mean

f_{i} = frequency of ith class

d_{i} = x_{i} – a = deviation of ith class

Î£f_{i} = n = Total number of observations

x_{i} = class mark = (upper class limit + lower class limit)/2

The value for a will be taken one among x_{i}â€™s with the assumption that the frequency of a class is centred at its mid-point, called its class mark.

Read more:

Relation Between Mean Median and Mode

## Assumed Mean Method Questions

If x_{i} and f_{i} are numerically large, the assumed mean method is preferred. Below are some examples of calculating the mean of a grouped data by this method.

**Example 1:**

The following table gives the information about the marks obtained by 100 students in an examination.

Class |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |

Frequency |
12 |
28 |
32 |
25 |
13 |

Find the mean marks of the students using assumed mean method.

Solution:

Class (CI) |
Frequency (f |
Class mark (x |
d |
f |

0-10 |
12 |
5 |
5 – 25 = – 20 |
-240 |

10-20 |
28 |
15 |
15 – 25 = – 10 |
-280 |

20-30 |
32 |
25 = a |
25-25 = 0 |
0 |

30-40 |
25 |
35 |
35-25 = 10 |
250 |

40-50 |
13 |
45 |
45-25 = 20 |
260 |

Total |
Î£f |
Î£f |

Assumed mean = a = 25

Mean of the data:

= 25 + (-10/ 100)

= 25 – 1/10

= (250-1)/10

= 249/10

=24.9

Hence, the mean marks of the students are 24.9.

**Example 2:**

The table below gives the information about the percentage distribution of female employees in a company of various branches and number of departments.

Percentage of female employees |
Number of departments |

5-15 |
1 |

15-25 |
2 |

25-35 |
4 |

35-45 |
4 |

45-55 |
7 |

55-65 |
11 |

65-70 |
6 |

Find the mean percentage of female employees by assumed mean method.

Solution:

Percentage of female employees (CI) |
Number of departments (f |
Class mark (x |
d |
f |

5-15 |
1 |
10 |
-30 |
-30 |

15-25 |
2 |
20 |
-20 |
-40 |

25-35 |
4 |
30 |
-10 |
-40 |

35-45 |
4 |
40 = a |
0 |
0 |

45-55 |
7 |
50 |
10 |
70 |

55-65 |
11 |
60 |
20 |
220 |

65-70 |
6 |
70 |
30 |
180 |

Total |
Î£f |
Î£f |

Assumed mean = a = 40

Mean = a+ (Î£f_{i}d_{i} /Î£f_{i})

=40+ (360/35)

= 40+(72/7)

= 40 + 10.28

=50.28 (approx)

Hence, the mean percentage of female employees is 50.28.

Thank you sir/ma for this, I learnt alot Im grateful. I have a question Is the assumed mean always centrally placed in the class mark?. Thank you.

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