# Assumed Mean Method

In statistics, the assumed mean method is used to calculate 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. 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 computations are performed.

## Assumed Mean Method Formula

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

Here,

a = assumed mean

f_{i} = frequency of ith class

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

Σf_{i} = n = Total number of observations

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

## Assumed Mean Method Examples

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

**Example 1:**

The following table gives information about the marks obtained by 110 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 the assumed mean method.

**Solution:**

Class (CI) | Frequency (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – a |
f_{i}d_{i} |

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 | Σf_{i} =110 |
Σf_{i}d_{i} = -10 |

Assumed mean = a = 25

Mean of the data:

= 25 + (-10/ 110)

= 25 -( 1/11)

= (275-1)/11

= 274/11

=24.9

Hence, the mean marks of the students are 24.9.

**Example 2:**

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

65-75 | 6 |

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

**Solution:**

Percentage of female employees (CI) | Number of departments (f_{i}) |
Class mark (x_{i}) |
d_{i} = x_{i} – a |
f_{i}d_{i} |

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 |

65-75 | 6 | 70 | 30 | 180 |

Total | Σf_{i} =35 |
Σf_{i}d_{i} = 360 |

Assumed mean = a = 40

Mean = a+ (Σf_{i}d_{i} /Σf_{i})

=40+ (360/35)

= 40+(72/7)

= 40 + 10.28

=50.28 (approx)

Hence, the mean percentage of female employees is 50.28.

### Practice Questions on Assumed Mean Method

Solve the following questions using the formula of assumed mean method.

1. Find the mean of the following data by assumed mean method.

CI | 20 – 60 | 60 – 100 | 100 – 150 | 150 – 250 | 250 – 350 | 350 – 4507 |

f | 7 | 5 | 16 | 12 | 2 | 3 |

2. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.

Runs scored | Number of batsmen |

3000 – 4000 | 4 |

4000 – 5000 | 18 |

5000 – 6000 | 9 |

6000 – 7000 | 7 |

7000 – 8000 | 6 |

8000 – 9000 | 3 |

9000 – 10000 | 1 |

10000 – 11000 | 1 |

Find the mean of the data.

3. Find the mean of the following data using the assumed mean method formula.

Marks | Number of students |

0 – 10 | 5 |

10 – 20 | 3 |

20 – 30 | 4 |

30 – 40 | 3 |

40 – 50 | 3 |

50 – 60 | 4 |

60 – 70 | 7 |

70 – 80 | 9 |

80 – 90 | 7 |

90 – 100 | 8 |