First of all, one needs to understand that this is another method of finding the Mean of a list of data. Sometimes the data is presented in a form which makes the calculation of the mean by the usual method impossible. Such a situation would involve a frequency table where the data is presented as an interval. For example, a question might be, “A person rolls a 1, 2 or 3 five times, then rolls a 4, 5 or 6 three times. What is the mean roll value?” Since we do not know exactly how many times each number has been rolled, we need another method to solve this problem. This would be an occasion to use an Assumed Mean. To solve the problem one chooses an arbitrary number which is close to the mean, in this case we could use 6 as the assumed value. Since it is an arbitrary choice, we could also use 5, or 2. It doesn’t matter.

Next we set up a table as shown below

Dice value

Frequency (f) Mid-Value (x) d = x – A fd A

1-3

5 2 -4 -20 6

4-6

3 5 -1 -3 8 -23 fd/f A + (fd/f) -2.875 3.125

We find the mid-value of each interval, and subtract the Assumed value (A) in this case 6, then we multiple that value (-4) times the frequency (5) to yeild -20. Continue down the table. In the end, we add up our frequencies (8) and f x d (-23), then divide fd by d. We then add this value to the assumed mean (6 + (-2.875)) and this is our mean 3.125

## Statistics – Assumed Mean Method

Like its, name in this method we assume a number as the mean of the given data.

Then we subtract this assumed mean from all class marks. Hence find d_{i}.

Let a is our assumed mean and x_{1}, x_{2}, x_{3} …….x_{n} are the class marks. Then

d1 = x_{1 }– a

d_{2} = x_{2} – a

d_{3} = x_{3} – a

————-

————-

d_{n} = x_{n} – a

Collectively we can write

d_{i} = x_{i} – a

After that we calculate

∑f_{i} d_{i} = f_{1} d_{1} + f_{2} d_{2} + f_{3} d_{3} + ——- + f_{n} d_{n}

Now

Since d_{i }= x_{i} – a

Therefore

This is the formula for calculating mean with assumed mean method. The method

will be more clear with following example** **

**Note : **You can assume any number as assumed mean (a) but the general choice is

any xi which lies in the middle of range of x_{i}.

My forthcoming post is on Correlation Coefficient Definition with example,and this topic get answers to math problems will give you more understanding about Math.

## Assumed mean-Example

Calculate the mean of following data with the help of assumed

mean method

C.I. |
10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 | 80-90 | 90-100 | 100-110 |

f_{i} |
4 | 5 | 4 | 4 | 3 | 4 | 16 | 13 | 11 | 16 |

**Solution:**

C.I. |
X_{i} |
f_{i} |
d = x_{i} – a |
f_{i}di |

10-20 | 15 | 4 | -40 | -160 |

20-30 | 25 | 5 | -30 | -150 |

30-40 | 35 | 4 | -20 | -80 |

40-50 | 45 | 4 | -10 | -40 |

50-60 | 55 ? a | 3 | 0 | 0 |

60-70 | 65 | 4 | 10 | 40 |

70-80 | 75 | 16 | 20 | 320 |

80-90 | 85 | 13 | 30 | 390 |

90-100 | 95 | 11 | 40 | 440 |

100-110 | 105 | 16 | 50 | 800 |

`sum` f_{i=80} |
`sum` f_{i}d_{i= 1560} |

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Let a = 55

Now

= 55 + (1560 / 80)

= 55 + 19.5

= **74.5**