## WBBSE Solutions For Class 10 Maths Statistics Chapter 2 Ogive

Ogive is the curved line or graph of a cumulative frequency distribution (Less or more than type).

**By the help of ogive, we can find the median of a frequency distribution. There are two types of ogives**

- Ogive of less than type cumulative frequency distribution
- Ogive of more than type cumulative frequency distribution.

**Construction of Ogive:**

We can construct two types of giving

- Ogive of less than type cumulative frequency distribution;
- Ogive of more than type cumulative frequency distribution.

**Construction of ogive of less than type cumulative frequency distribution:**

In order to construct the ogive of less than type cumulative frequency distribution, we first, determine the scale of the graph paper, i.e., we fixed up a certain scale of both the axes of the graph paper.

**WBBSE Solutions for Class 10 Maths**

For cumulative frequency, we denote the upper boundaries of each class interval along the horizontal line, i.e. along the x-axis, and their corresponding cumulative frequency along the- vertical line, i.e., along the y-axis.

Later on, the points thus obtained are plotted in the graph paper. Now, the points are joined by a scale or in open hands to get a curved

line.

This curved line is called ogive. Observe the following examples:

**Example:**

Class intervals |
100-20 |
120-140 |
140-160 |
160-180 |
180-200 |

Frequencies |
12 | 14 | 8 | 6 | 10 |

Find the median of the above frequency distribution by constructing ogives of less than and more than type cumulative frequencies along the same axes.

**Solution:** At first, let us construct a less-than-type and a more-than-type cumulative frequency distribution table of the above data.

Class interval |
Classes |
More than type cumulative frequencies |
Classes |
Less than type cumulative frequencies |

100-120 | 100 or more than 100 | 50 | Less than 120 | 12 |

120-140 | 120 or more than 120 | 38 | Less than 140 | 26 |

140-160 | 140 or more than 140 | 24 | Less than 160 | 34 |

160-180 | 160 or more than 160 | 16 | Less than 180 | 40 |

180-200 | 180 or more than 180 | 10 | Less than 200 | 50 |

**Selection of Scale**:

Let one side of each smallest square = 10 units along the x-axis and one side of each smallest square = 2 units along the y-axis.

**Plotting of points:**

For ogive of more than type cumulative frequency distribution, let us plot the points ( 1 00, 50), (120,38), (140, 24), (160, 16), and (180, 10).

But according to the above scale (10, 25), (12, 19), (14, 12), (16, 8) and (1 8, 5). Later on, let us plot the points on the same graph paper.

Then, joining the points by free-hand of by a scale, we get an ogive of more than type cumulative frequency.

For ogive of less than type cumulative frequency distribution, let us plot the points ( 1 20, 1 2), (140, 26),(160, 34), (180, 40), and (200, 50).

But according to the above scale the points will be transferred to the new points (12, 6), (14, 13), (16, 17), (18, 20), and (20, 25). Then let us plot the points on the same graph paper.

We thus get an Ogive of less than type cumulative frequency distribution.

We see from the graph paper that the two ogives intersect each other at P.

Now, let us draw a perpendicular PM from the point P on the x-axis. Taking measure of the perpendicular PM by a scale we get, PM = 13.857 (approx.)

But according to the scale, 13.857 square = 13.857 x 10 units (approx.) = 138.57 units (approx.)

The median of the data = 138-57 units (approx.)

We can thus determine the median of any given data by constructing ogives.

**Now, we prove this determination of this median by formula directly:**

**Formula:** Median = \(l+\left[\frac{\frac{n}{2}-c f}{f}\right] \times h\)

Here, 10 = 140, n = 50, \(\frac{n}{2}\) = 25, cf = 26, f = 14, h = 20

∴ Median = 140 + \(\frac{25-26}{14}\) x 20 = 140 – 1.4285 = 138.57

∴ The required median = 138.57 units(approx).