PUMPA - THE SMART LEARNING APP

Take a 10 minutes test to understand your learning levels and get personalised training plan!

Download now on Google PlayLet us understand the concept of

**median**and**mode**using an example.Priya and Susan are friends having a discussion among themselves that who was the least in spending money. Their last spending for the previous \(5\) months are:

Priya's spending (in rupees) | \(100\) | \(250\) | \(150\) | \(220\) | \(150\) |

Susan's spending (in rupees) | \(90\) | \(375\) | \(90\) | \(90\) | \(250\) |

They both determined their average spending's as:

Priya's average \(= \frac{100 + 250 + 150 + 220 + 150}{5}\) \(= \frac{870}{5}\) \(= ₹174\)

Susan's average \(= \frac{95 + 375 + 90 + 90 + 250}{5}\) \(= \frac{895}{5}\) \(= ₹179\)

Since Priya's average spending is the lowest, Priya proved to have spent the least in spending money, but Susan cannot agree with it. So, Susan arranged both their spendings in ascending order. Thus, the table looks like this:

Priya's spending (in rupees) | \(100\) | \(150\) | \(150\) | \(220\) | \(250\) |

Susan's spending (in rupees) | \(90\) | \(90\) | \(90\) | \(250\) | \(375\) |

From the above table, Susan showed Priya that her middle value (\(90\)) is the lowest, and hence, her spending was the least. Since Priya isn't satisfied, Susan showed that she repeatedly spend only \(₹90\) for \(3\) months which were the least and therefore, Susan claimed to have the least spending.

To settle their dispute, we arrive at the methods of solving the central measures.

The first step of finding the average spending is the mean. And, second, Susan showed the middle value, which is the median. And finally, the most often occurred number in spending is the mode.

Median

Median is defined as the middle value which exactly divides the given set of observations into two equal parts.

- If the number of observations (\(n\)) in the data set is odd, then the median can be determined using the formula \((\frac{n+1}{2})^{th}\) observation.
- If the number of observations (\(n\)) in the data set is even, then the median is the mean of the values \((\frac{n}{2})^{th}\) and \((\frac{n}{2}+1)^{th}\) observations.

Mode

Mode is defined as the number which most frequently occurs in the given set of data. That is, the observation having the maximum number of frequency is called mode.

Using these \(3\) measures of central tendency, let us solve the dispute between Susan and Priya.

We have already determined the mean of these two persons. Now, we shall find the median.

Since the number of observations for both persons is odd\((5)\), the median can be determined using the formula, \(\frac{n+1}{2}\).

Substituting \(n = 5\), we have:

Median \(= \frac{5 + 1}{2} = \frac{6}{2}\) \(= 3\)

That is, the median for Priya is \(150\), and the median for Susan is \(90\).

Similarly, the mode for Priya's spending is \(150\), and the mode for Susan's spending is \(90\).

Tabulating the \(3\) values of central measures of tendency, we have:

Central measures | Priya | Susan |

Mean | \(174\) | \(179\) |

Median | \(150\) | \(90\) |

Mode | \(150\) | \(90\) |

From these values, we can understand these comparison does not help in identifying the person who spends the least money in the past \(5\) months. Hence we need some more data to conclude this, which we will learn in the higher classes.

Important!

A distribution having only \(1\) mode is called unimodal.

A distribution having \(2\) modes is called bimodal.

A distribution having \(3\) modes is called trimodal.

A distribution having more than \(3\) modes is called multimodal.