** MEDIAN, **ADVANTAGES AND DISADVANTAGES OF MEDIAN. Definition of median: The median is defined as an average, which is the middle value when figures are arranged in order of magnitude. In an even distribution, the median is the average of the two middle numbers.

I other words, the median of a distribution is the middle value when the observations are arranged in order of magnitude starting with either the smallest or the largest number. The median is therefore the value of the middle item.

## How to calculate the median

- When the numbers involved are odd numbers, the median will be the middle number.

Example I

Find the median of the following sets of values 2,8,11,13,15,6,9,20,7

**Solution**

**Formula: ** (n + 1)

** Median ** 2

Where n is the number of items (observation). This formula is applicable only when n is odd.

**Step 1** First arrange the numbers in ascending or descending magnitude

**In ascending magnitude, we have**2, 6, 7, 8, 9, 11, 13, 15, 20.

**Count the numbers of values involved**

There are nine (9) numbers involved (odd number)

Since there are 9 numbers involved, the middle umber is the 5^{th} number and the 5^{th} number is 9. Therefore, the median = 9

Alternatively, the formula can be used

**Median = **(n + 1)

** ** 2

= 9 + 1 = 10 = 5

2 2

The 5^{th} number in ascending magnitude is 9

Therefore the median = 9

**In ascending magnitude, we have**20, 15, 13, 11, 9, 8, 7, 6, 2

Since there are nine (9) numbers involved, the 5^{th} number, which is the middle one (9), is

SD

Therefore the median = 9

- When even numbers are representing the number of events in data, the two middle values are taken; add them and divide then divide by two. The median will be the arithmetic mean of the two middle numbers.

Example 2

Find the median of the following numbers: 2, 8, 12, 8, 10, 14, 18, 5

Solution

Step 1: First arrange the numbers in ascending or descending magnitude. In ascending magnitude, we have: 5, 8, 8, 10, 12, 14, 18, 22

Step II: Count the number of values involved

There are eight (8) values (even numbers)

Step III: Since there are eight (8) numbers involved, the middle will be the 4^{th} and the 5^{th} numbers, which are 10 and 12.

Step IV: To get the median, then add 10 + 1 together and divide them by 2

i.e. 10 + 12 =22 =11

2 2

The median = 11

## Alternatively, the formula can be used for median calculation

Median >>>>>>>>>>>>>>>>>>>

Therefore the median is found between 10 and 12. The middle between 10 and 12 is 11

Therefore, median = 11

- When a group data is involved, cumulative frequency is used

This is used when items or values are large and arranging them in ascending order may not work. The formula will now be:

FORMULA >>>>>>>>>>>>>>>>>>>>>

There N is the summation of all the frequency and this is the terminal cumulative frequency.

Example 3

Use the information in table 2.13 and calculate the median age of some SSIII students.

Table 2.13: Age distribution of SSIII students

Age (yrs) | 6 | 8 | 10 | 11 | 12 | 13 | 14 |

Frequency | 5 | 10 | 3 | 8 | 7 | 10 | 8 |

**Solution**

Cumulative frequency table (table 2.14) is prepared for distribution.

Table 2.14: Cumulative frequency for age distribution of SIII students

Age (yrs) | 6 | 8 | 10 | 11 | 12 | 13 | 14 |

No. of students (frequency) | 5 | 10 | 3 | 8 | 7 | 10 | 8 |

Cumulative frequency | 5 | 15 | 18 | 26 | 33 | 43 | 51 |

From table 2.14, there are 51 members as indicated by the terminal (last) cumulative frequency. Since the members are **odd **(51) the median age will be (N+1)th member

2

= 52

2

= 26^{th} member

The 26^{th} falls within the cumulative under the age of 11years in the table above. Therefore, the median age = 11years

**Example 4 **

The data in table 2.15 represents the marks scored by Government students in NECO examinations. Calculate the median score.

Table 2.15: Marks scored by government students in NECO examinations.

Marks % | 12 | 18 | 24 | 30 | 36 | 40 | 48 |

Frequency | 6 | 1 | 10 | 8 | 12 | 3 | 4 |

Solution

A cumulative frequency table (table 2.16) is prepared for the distribution .

Table 2.16: Cumulative frequency table for marks scored by Government Students in NECO examination

Marks % | 12 | 18 | 24 | 30 | 36 | 40 | 48 |

Cumulative Frequency | 6 | 1 | 10 | 8 | 12 | 3 | 4 |

Frequency | 6 | 7 | 17 | 25 | 37 | 40 | 44 |

From the table (table 2.16), there are 44 members as indicated by the terminal (last) cumulative frequency.

Since this 44 is even, the median score will be:

Median score = N th + N + 1 th

2 2

= 44 th + 44 + 1 th

2 2

= 22nd + 23^{rd}

3

The 22nd member is 30marks

The 23rd member is 30marks

Median score = 30 + 30 = 60 = 30marks

2 2

Median score = 30

### Advantages of median

- Computation in median is very easy
- Median is not affected by extremes of values
- Advantages of the median it is very easy to understand
- It can be obtained by graphic form
- The median is easy to determine by mere observation
- Another advantage of median is that It does not involve serious calculations

#### Disadvantages of median

WEED AND THEIR BOTANICAL NAMES

1. ENVIRONMENTAL FACTORS AFFECTING AGRICULTURAL PRODUCTION

2. DISEASES

3. 52. SOIL MICRO-ORGANISMS

4. ORGANIC MANURING

5. FARM YARD MANURE

6. HUMUS

7. COMPOST

8. CROP ROTATION

9. GRAZING AND OVER GRAZING

- IRRIGATION AND DRAINAGE
- IRRIGATION SYSTEMS
- ORGANIC MANURING
- FARM YARD MANURE
- HUMUS
- COMPOST
- CROP ROTATION

- IRRIGATION AND DRAINAGE
- IRRIGATION SYSTEMS
- INCUBATORS
- MILKING MACHINE
- SIMPLE FARM TOOLS
- AGRICULTURAL MECHANIZATION
- THE CONCEPT OF MECHANIZATION
- PROBLEMS OF MECHANIZATION
- SURVEYING AND PLANNING OF FARMSTEAD
- IMPORTANCE OF FARM SURVEY
- SURVEY EQUIPMENT
- PRINCIPLES OF FARM OUTLAY
- SUMMARY OF FARM SURVEYING
- CROP HUSBANDRY PRACTICES

- Difficulties come up when large values are involved
- One of the disadvantages of median is the re-arrangement of numbers involve a difficult task
- disadvantages of median is that It may not be needed for further statistical calculations
- It tends to ignore extreme values