## Median

**Median **of an observation is the value of the variable which divides it into two equal parts. It is the value which is

- If there are odd number of observations, the value in the middle of the ordered list.
- If there are even number of observations, it is the arithmetic mean of values in the middle of the ordered list.

The median is a positional average. Sometimes, in case of even number of observations, instead of arithmetic mean of values in the middle, one of the two is chosen randomly.

### Steps to Calculate Median

- Find N/2, where N is the total number of observations.
- Sort the observations and tabulate it. Add a third column to calculate the cumulative (total) frequency.
- Spot the
*x*with cumulative frequency just greater than N/2. That will be your median.

**Example 7**: Find the median of 2, 3, 3, 5, 6, 6, 6**Solution**: Note that N = 7, hence, N/2 = 3.5. So, the 4th observation, x =5 is the median.

**Example 8**: Find the median of 1, 11, 11, 13, 16, 17.**Solution**: Note that N = 6. The values in centre are, 11 and 13. Then by the rule, we can choose 11 or 13 as the median. Or, we can take their average, (11+13)/2 = 12 as their median as well.

**Example 9**: Pick any 5 numbers. Find its (arithmetic) mean and median.

**Example 10**: Obtain the median for the following frequency distribution:

x: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

f: | 8 | 10 | 11 | 16 | 20 | 25 | 15 | 9 | 6 |

**Solution**: Here, the value of *x*’s are already sorted.

x | f | Cumulative Frequency |

1 | 8 | 8 |

2 | 10 | 8+10 = 18 |

3 | 11 | 18+11= 29 |

4 | 16 | 29+16= 45 |

5 | 20 | 45+20= 65 |

6 | 25 | 65+25= 90 |

7 | 15 | 90+15= 105 |

8 | 9 | 105+9= 114 |

9 | 6 | 114+6= 120 |

N = 120 |

Hence, N = 120 and N/2 = 60.

Observe the cumulative frequency column. The value just greater than 60 in it is 65. Corresponding value to 65 of *x* is 5.

Hence, median is 5.

In the case of continuous frequency distribution, the class corresponding to the *cf*. just greater than NI2 is called the median class and the value of median is obtained by the following formula:

Median =

where *l* is the lower limit of the median class.

*f* is the frequency of the median class.

*h* is the magnitude of median class

‘*c*’ is the *cumulative frequency* of the class preceding the median class.

and N = **∑ ***f*

**Example 11**: Find the median wage of the following distribution:

Wages (in Rs.): | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |

No. of labourers: | 3 | 5 | 20 | 10 | 5 |

**Solution**:

Wages (x, in Rs.) | No. of labourers (f) | Cumulative Frequency |

20-30 | 3 | 3 |

30-40 | 5 | 3+5 = 8 |

40-50 | 20 | 8+20 = 28 |

50-60 | 10 | 28+10 = 38 |

60-70 | 5 | 38+5 = 43 |

N = 43 |

Now, N/2 = 43/2 = 21.5.

Cumulative frequency just greater than 21.5 is 28 and corresponding class is 40-50.

Thus, the median class is 40-50. Thus,

*l = 40, f = 28, h = 10, c = 8, N = 43.*

Using the formula,

Median =

**Example 12**: Find the median of the following distribution:

Wages (in Rs.): | 10-30 | 30-50 | 50-70 | 70-90 |

No. of labourers: | 4 | 6 | 18 | 12 |

(Try it yourself!)

### Merits and Demerits of Median

Requisite | Merit | Demerit | Notes |

It is rigidly defined. | ✓ | ||

It is easy to understand and calculate. | ✓ | …and in some cases, it can be calculated by just looking at the distribution! | |

It is not affected much by extreme observations. | ✓ | ||

It is not suitable for further mathematical treatment. | ✓ | ||

It is not based on all observations | ✓ | Consider the distribution 10, 14, 20, 32, 35. Even if we replace 10 and 14 by two numbers less than 20, median will not change. | |

In comparison to mean, it is affected by fluctuation in sampling. | ✓ |

## Mode

**Mode **is the value which occurs most frequently in a set of observations, and around which other values tend to be around. The mode of a distribution is the value of x corresponding to maximum frequency. A distribution may have two modes (bi-modal) or more than two (multimodal).

**Example 13**: Find the mode of the following distribution:

x: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

f: | 4 | 10 | 11 | 16 | 20 | 25 | 15 | 9 | 6 |

**Solution**: 25 has the highest frequency. Hence the mode is the corresponding *x* value, 6.

**Example 14**: Find the mode of the following distribution:

x: | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |

f: | 4 | 10 | 11 | 16 | 20 | 25 | 15 | 9 | 6 |

**Solution**: (Try it yourself!)

In case of class-intervals, suppose *x _{k}-x_{k+1} *is the modal class with highest frequency

*f*then

_{k},where *l* is the lower limit of the modal class.

**Example 15**: Find the mode of the following distribution:

Class Interval: | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | 70-80 |

Frequency: | 5 | 8 | 7 | 12 | 28 | 20 | 10 | 10 |

**Solution**: Here, maximum frequency is 28. Thus the class 40-50 is the modal class.

Then, *l = 40, h = 10, f _{k} = 28, f_{k-1} = 12, f_{k+1} = 20.*

Mode =

**Example 16**: Find the mode of the following distribution:

Class Interval: | 5-10 | 10-15 | 15-20 | 20-25 | 25-30 |

Frequency: | 5 | 8 | 17 | 12 | 8 |

**So****lution**: (Try it yourself!)

Sometimes, mode can be approximated from mean, using the formula Mode = 3 Median – 2 Mean.

Requisite | Merit | Demerit | Notes |

It is ill-defined. | | ✓ | Mode of a distribution need not be unique. |

Easy to understand and calculate. | ✓ | …and in some cases, simply by inspection. | |

We can find mode of open-ended classes | ✓ | ||

It is not at all affected by extreme observations. | ✓ | ||

It is not suitable for further mathematical treatment. | ✓ | ||

It is not based on all observations | ✓ | ||

In comparison to mean, it is affected by fluctuation in sampling. | ✓ |

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