The **assumed mean method** is a technique used in statistics to calculate the **arithmetic mean**. It is a useful *shortcut method* to calculate the mean from a set of data.

Index

## Assumed Mean Method Formula

Here, we present the assumed mean method formula for different kinds of data.

### Discrete Data

Let us consider a collection of \(\boldsymbol{N}\) values whose mean we have to find. Each value is denoted by \(x_i\). We make a guess and choose an** approximate mean**, \(x_0\), roughly in the middle of the data. This is called the **assumed mean.**

Then, we calculate **deviations** from the assumed mean, defined as:

\(d_i = x_i – x_0\)

Now, we add up these deviations for all values in the data.

\(A = \sum_{i=1}^N d_i\)

This is divided by the** total number of observations** \(N\). This value \(D\) is the difference between the actual mean and the assumed mean.

\(D = \frac{A}{N}\)

Finally, the actual mean is obtained as follows.

\(\bar{x} = x_0 + D\)

### Grouped Data

The process is very similar in grouped data. The only major differences are as follows:

- For each class, the assumed mean is subtracted from the
**class mark**(midpoint) of class interval. - The individual deviations \(d_i\) is multiplied by the
**frequency**\(f_i\) of the class intervals.

Thus, the assumed mean formulae become,

\(d_i = x_i – x_0\)

Where, \(x_i\) are **class marks.**

\(A = \sum_{i=1}^N f_i d_i\)

Where, \(f_i\) is **class frequency.**

\(D = \frac{A}{N}\)

Finally,

\(\bar{x} = x_0 + D\)

## Applications of Assumed Mean Method

This method can be used to make the calculation of mean *much faster by hand*. If the assumed mean is chosen well, the deviations tend to be small and almost cancel out, making the addition easier.

The process used for grouped data also speeds up the procedure very much, providing accurate results.

## Example Questions

**Question 1.** Find the mean of this set of values by assumed mean method.

91, 48, 9, 37, 6, 42, 23, 45, 63, 84, 88, 29, 28, 10, 8

**Solution.** Let us assume mean \(x_0 = 40\). Then the deviations \(x_i\) from mean become,

\(51, 8, -31, -3, -34, 2, -17, 5, 23, 44, 48, -11, -12, -30, -32\)

Now, we have their sum \(A = 11\).

Now, the difference \(D\) between actual mean and assumed mean is

\(D = \frac{A}{N}\)

So, \(D = \frac{11}{15} = 0.733\).

Finally, the actual mean is given by,

\(\bar{x} = x_0 + D = 40 + 0.733 = 40.733\)

So the mean of the above data is \(\bar{x} = 40.733\). *A good choice of assumed mean simplified the process significantly. *

**Question 2.** Find the mean of the following data, using classes of width 10.

66,75,79,56,61,77,92,75,78,51,82,85,59,66,91,98,90,73,71,83,85,91,77,70

**Solution.** Let us group the data into the following classes: \(50-60, 60-70, 70-80, 80-90, 90-100\). Then we have can arrange the required values in a table as follows.* Let us assume a mean of 75.*

Class Interval | Class Mark | Frequency \(\boldsymbol{f_i}\) | Deviation \(\boldsymbol{d_i}\) | \(f_i \times d_i\) |
---|---|---|---|---|

50 – 60 | 55 | 3 | -20 | -60 |

60 – 70 | 65 | 3 | -10 | -30 |

70 – 80 | 75 | 9 | 0 | 0 |

80 – 90 | 85 | 4 | +10 | +40 |

90 – 100 | 95 | 5 | +20 | +100 |

Sum | \(N = 24\) | \(A = +50\) |

Thus, we get \(D = \frac{A}{N} = \frac{50}{24}\). So, \(D \approx 2.1\).

We then have **the actual mean** as follows.

\begin{align}

\bar{x} & = x_0 + D \\

& = 75 + \frac{50}{24} \\

& \approx 75 + 2.1 \\

\bar{x} & = 77.1

\end{align}

\)

## FAQs

**How do you calculate the assumed mean?**We calculate the mean by the assumed mean method for a series of \(N\) values \(x_i\) as follows.

1. Assume a mean \(x_0\).

2. Calculate **deviations** \(d_i = x_i – x_0\) from assumed mean, for each \(x_i\).

3. Sum the deviations for all values. \(A = \sum_{i=1}^N d_i\).

4. Calculate \(D = \frac{A}{N}\).

Find **actual mean** \(\bar{x}\) as follows: \(\bar{x} = x_0 + D\).

**How can you find the**mean for grouped data by the**assumed mean method?**The process is the same as the method used for discrete data, with the only differences being:

1. The values \(x_i\) become the **class-marks** for each class interval.

2. The deviations \(d_i\) are* multiplied* by the **frequency** \(f_i\) of each class interval before being added.

In short,

\(\bar{x} = x_0 + \frac{\sum f_i d_i}{N}\)

Where, \(N = \sum f_i\)

**What is the difference between actual mean and assumed mean?**The** actual mean** \(\bar{x}\) is the real arithmetic mean of the given data.

The **assumed mean** \(x_0\) is a guess of the *approximate value for the mean.* It is used in the **assumed mean method** to simplify calculations.

**How can you find standard deviation by the assumed mean method?**We define, \(B = \sum_{i=1}^N {d_i}^2\).

Then, we have standard deviation \(\sigma\) as,

\(\sigma = \sqrt{\frac{B-ND^2}{N}}\)

Where, \(N\) is **number of observations**, \(D\) is *differenc*e between \(\bar{x}\) and \(x_0\) as defined above.