UPSKILL MATH PLUS

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Learn moreThe median of an ungrouped frequency distribution can be determined using the following steps.

**1**. Arrange the given data in ascending or descending order.

**2**. Find the cumulative frequency distribution and denote \(N\) as the total frequency.

**3**. If \(N\) is

**odd**, then median \(= \left(\frac{N + 1}{2} \right)^{th}\) term.

**4**. If \(N\) is

**even**, then median \(= \left(\frac{(\frac{N}{2}^{th}) \text{observation} + (\frac{N}{2}+1)^{th} \text{observation}}{2} \right)\)

Example:

Find the median of the following data.

Marks | \(30\) | \(35\) | \(60\) | \(92\) | \(85\) | \(75\) |

Number of students | \(1\) | \(4\) | \(6\) | \(2\) | \(10\) | \(7\) |

**Solution**:

Let us arrange the marks in ascending order and find the cumulative frequency.

Marks | Frequency(\(f\)) | Cumulative frequency(\(cf\)) |

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

\(35\) | \(4\) | \(1 + 4 = 5\) |

\(60\) | \(6\) | \(5 + 6 = 11\) |

\(75\) | \(7\) | \(11 + 7 = 18\) |

\(85\) | \(10\) | \(18 + 10 = 28\) |

\(92\) | \(2\) | \(28 + 2 = 30\) |

Therefore, the total frequency \(N = 30\).

Since \(N\) is even, the median can be determined using the formula \(\left(\frac{(\frac{N}{2}^{th}) \text{observation} + (\frac{N}{2}+1)^{th} \text{observation}}{2} \right)\)

Substituting the known values, we get:

Median \(= \left(\frac{(\frac{30}{2}^{th}) \text{observation} + (\frac{30}{2}+1)^{th} \text{observation}}{2} \right)\)

\(= \left(\frac{15^{th} \text{observation} + 16^{th} \text{observation}}{2} \right)\)

If the marks of the students are arranged in ascending order, then the \(15^{th}\) and \(16^{th}\) terms would be in the middle value. From the cumulative frequency distribution, let us select the marks near the \(15^{th}\) and \(16^{th}\) terms. Thus, the marks would be \(75\).

Substituting these values in the median, we have:

Median \(= \frac{75+75}{2}\) \(=75\)

Therefore, the median of the given data is \(75\).