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Download now on Google PlayThe mean of the ungrouped frequency distribution can be determined using the formula:

\(\overline X = \frac{f_1 x_1 + f_2 x_2 + ... + f_n x_n}{f_1 + f_2 + ... + f_n}\) \(= \frac{\sum_{i=1}^{n} f_i x_i}{\sum_{i=1}^{n} f_i}\) \(= \frac{\sum fx}{\sum f}\)

Example:

The marks scored by \(20\) students in mathematics are \(100\), \(65\), \(89\), \(55\), \(40\), \(96\), \(65\), \(35\), \(35\), \(100\), \(100\), \(100\), \(55\), \(65\), \(96\), \(89\), \(77\), \(63\), \(77\), \(30\).

**Solution**:

In the previous sessions, we have learnt how to find the average using the arithmetic mean method and assumed mean method.

Now, we shall learn the \(3^{rd}\) method, which is an ungrouped frequency distribution.

Here, in the data, we can find \(100\) has occured \(4\) times (i.e. frequency is \(4\)), \(65\) has frequency of \(3\) and so on. Then, the frequency distribution table looks like this:

Marks\(x\) | Number of students\(f\) |

\(30\) | \(1\) |

\(35\) | \(2\) |

\(40\) | \(1\) |

\(55\) | \(2\) |

\(63\) | \(1\) |

\(65\) | \(3\) |

\(77\) | \(2\) |

\(89\) | \(2\) |

\(96\) | \(2\) |

\(100\) | \(4\) |

We know the formula to find the mean of ungrouped frequency distribution is \(\overline X = \frac{\sum fx}{\sum f}\)

To find the value of \(fx\), multiply the value of \(x\) and \(f\) of each entry.

Consider for the mark \(30\). That is, \(30 \times 1 = 30\)

Similarly, for the mark \(35\), we have \(35 \times 2 = 70\) and so on.

Tabulating these values, we get:

Marks\(x\) | Frequency\(f\) | \(fx\) |

\(30\) | \(1\) | \(30\) |

\(35\) | \(2\) | \(70\) |

\(40\) | \(1\) | \(40\) |

\(55\) | \(2\) | \(110\) |

\(63\) | \(1\) | \(63\) |

\(65\) | \(3\) | \(195\) |

\(77\) | \(2\) | \(154\) |

\(89\) | \(2\) | \(178\) |

\(96\) | \(2\) | \(192\) |

\(100\) | \(4\) | \(400\) |

Total | \(\sum f = 20\) | \(\sum fx = 1432\) |

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

Mean \(\overline X = \frac{1432}{20}\) \(= 71.6\)

Therefore, the mean of the given data is \(71.6\).

We can also find the mean of the ungrouped frequency distribution using the assumed mean method.

Example:

Consider the above example. Let the assumed mean be \(A = 65\).

Then the new frequency distribution table is given by:

Marks\(x\) | Deviation \(d = x - A\) | Frequency\(f\) | \(fd\) |

\(30\) | \(30 - 65 = - 35\) | \(1\) | \(-35\) |

\(35\) | \(35 - 65 = - 30\) | \(2\) | \(-60\) |

\(40\) | \(40 - 65 = - 25\) | \(1\) | \(-25\) |

\(55\) | \(55 - 65 = - 10\) | \(2\) | \(-20\) |

\(63\) | \(63 - 65 = -2\) | \(1\) | \(-2\) |

\(65\) | \(65 - 65 = 0\) | \(3\) | \(0\) |

\(77\) | \(77 - 65 = 12\) | \(2\) | \(24\) |

\(89\) | \(89 - 65 = 24\) | \(2\) | \(48\) |

\(96\) | \(96 - 65 = 31\) | \(2\) | \(62\) |

\(100\) | \(100 - 65 = 35\) | \(4\) | \(140\) |

Total | \(\sum f = 20\) | \(\sum fd = 132\) |

Arithmetic mean \(=\) Assumed mean \(+\) Average of sum of deviations

Arithmetic mean \(= 65 + \frac{132}{20}\) \(= 65 + 6.6\) \(= 71.6\)

Therefore, the mean is \(71.6\).