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Consider if the data is very large and finding the products of the observations and then adding them becomes tedious and may result in errors. Let us use the assumed mean method to find the mean of grouped frequency to avoid such complications.

**Steps**:

**1**. Calculate the midpoint of the class interval and name it as \(x_i\).

**2**. From the data of \(x_i\), choose any value(preferably in the middle) as the assumed mean(\(a\)).

**3**. Determine the deviation \(d=x-a\) for each of the classes.

**4**. Multiply the deviation and frequency of each class interval and name it \(f_id_i\).

**5**. Find the values \(\sum f_id_i\) and \(\sum f_i\).

**6**. Calculate the mean by applying the formula \(\overline X = a + \frac{\sum f_id_i}{\sum f_i}\)

Example:

Find the mean of the following frequency distribution:

Class interval | \(10 - 20\) | \(20 - 30\) | \(30 - 40\) | \(40 - 50\) | \(50 - 60\) | \(60 - 70\) | \(70 - 80\) |

Frequency | \(23\) | \(15\) | \(10\) | \(28\) | \(5\) | \(7\) | \(11\) |

**Solution**:

Class interval | Frequency(\(f_i\)) | Midpoint(\(x_i\)) | Deviation\(d_i = x - 45\) | \(f_id_i\) |

\(10 - 20\) | \(23\) | \(15\) | \(-30\) | \(-690\) |

\(20 - 30\) | \(15\) | \(25\) | \(-20\) | \(-300\) |

\(30 - 40\) | \(10\) | \(35\) | \(-10\) | \(-100\) |

\(40 - 50\) | \(28\) | \(45\) | \(0\) | \(0\) |

\(50 - 60\) | \(5\) | \(55\) | \(50\) | \(250\) |

\(60 - 70\) | \(8\) | \(65\) | \(20\) | \(160\) |

\(70 - 80\) | \(11\) | \(75\) | \(30\) | \(330\) |

Total | \(\sum f_i = 100\) | \(\sum f_id_i = -350\) |

We know that the mean of grouped data using the assumed mean method can be determined using the formula \(\overline X = a + \frac{\sum f_id_i}{\sum f_i}\)

Substituting the known values in the above formula, we get:

\(\overline X = 45 + (\frac{-350}{100})\)

\(\overline X = 45 - 3.5\)

\(\overline X = 41.5\)

Therefore, the mean of the given data is \(41.5\).