# Ex.14.4 Q2 STATISTICS Solution - NCERT Maths Class 10

## Question

During the medical check-up of \(35\) students of a class, their weights were recorded as follows:

Weight in (Kg) |
Number of students |

Less than \(38\) | \(0\) |

Less than \(40\) | \(3\) |

Less than \(42\) | \(5\) |

Less than \(44\) | \(9\) |

Less than \(46\) | \(14\) |

Less than \(48\) | \(28\) |

Less than \(50\) | \(32\) |

Less than \(52\) | \(35\) |

Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.

## Text Solution

**What is known?**

The weight of \(35\) students of a class.

**What is unknown?**

The less than type ogive and median weight.

**Reasoning :**

The representation of cumulative frequency distribution graphically is known as a cumulative frequency curve, or an ogive.

**Steps:**

The given cumulative frequency distributions of less than type are

Weight in (Kg) |
Number of students |

Less than \(38\) | \(0\) |

Less than \(40\) | \(3\) |

Less than \(42\) | \(5\) |

Less than \(44\) | \(9\) |

Less than \(46\) | \(14\) |

Less than \(48\) | \(28\) |

Less than \(50\) | \(32\) |

Less than 52 | \(35\) |

Taking upper class limits on *\(x\)*-axis and their respective cumulative frequencies on *\(y\)-*axis, its ogive can be drawn as follows.

Here, \(n = 35\) \(\Rightarrow \frac n{2}=17.5\)

Mark the point \('A'\) whose ordinate is \(17.5\) and its *\(x\)*-coordinate is \(46.5.\) Therefore, median of this data is \(46.5.\)

It can be observed that the difference between two consecutive upper-class limits is \(2.\) The class marks with their respective frequencies are obtained as below

Weight (in Kg) |
Frequency |
Cumulative Frequency |

Less than \(38\) | 0 | \(0\) |

\(38\) - \(40\) | \(3 - 0 = 3\) | \(3\) |

\(40\) - \(42\) | \(5 - 3 = 2\) | \(5\) |

\(42\) - \(44\) | \(9 - 5 = 4\) | \(9\) |

\(44\) - \(46\) | \(14 - 9 = 5\) | \(14\) |

\(46\) - \(48\) | \(28 - 14 = 14\) | \(28\) |

\(48\) - \(50\) | \(32 - 28 = 4\) | \(32\) |

\(50\) - \(52\) | \(35 - 32 = 3\) | \(35\) |

Total |
\(n=35\) |

Cumulative frequency \((cf)\) just greater than \(17.5\) is \(28\), beloging to class \(46-48.\) Therefore,median class \(=46-48\)

Class size (\(h\)) \(= 2\)

Lower class limit of median class (\(l\))\(=46\)

Frequency of median class (\(f\))\(=14\)

Cumulative frequency of class preceding median class (\(cf\)) \(=14\)

\[\begin{align}\text{Median} &= l + \left( {\frac{{\frac{n}{2} - cf}}{f}} \right) \times h\\ & = 46 + \left( {\frac{{17.5 - 14}}{{14}}} \right) \times 2\\ &= 46 + \frac{{3.5}}{7}\\ &= 46.5 \end{align}\]

Therefore, median of this data is \(46.5.\)

Hence, the value of median is verified.