## FREQUENCY TABLE:

Formation of frequency table for an ungrouped data

**Example:**

Consider the following data

16, 15, 20, 28, 40, 35, 36, 15, 16, 16, 37, 46, 45, 40, 28, 36, 16, 15, 40, 20, 28, 16, 15, 40, 20, 28, 16, 15 40, 28, 36, 37, 46, 40, 16

**Solution:**

The frequency table is given below

**Formation of frequency table for a grouped data**

**Example:**

The marks obtained by 40 students in English test with maximum marks of 100 are given as follows

50, 90, 36, 42, 10, 8, 52, 61, 75, 82, 93, 95, 47, 35, 65, 51, 48, 83, 66, 98, 26, 13, 12, 58, 86, 71, 72, 25, 31, 59, 53, 67, 81, 79, 89, 40, 11, 9, 39, 91.

Prepare a frequency table for the above data using class interval

**Solution:**

Total number of values= 40

Range= Highest value- Lowest value

Let us divide the given data into 10 classes

Length of class interval= Range

————————-

No. of class interval

= 90 = 9

——

10

The frequency table of the marks obtained by 40 students in English test is as follows

Drawing Histogram and Frequency Polygon for the grouped data

The statistical data can be represented by means of geometrical figures which are called “Graphs”. The graphical representation of data consumes less time and also its easy to understand.

Now we going to see two types of diagrams

(i) Histogram

(ii) Frequency Polygon

**Histogram:**

A two dimensional graphical representation of a continuous frequency distribution is called a Histogram

In Histogram the bars are placed continuously side by side with no gaps between adjacent bars. In histogram rectangles are erected on the class interval of the distribution. The areas of rectangle are proportional to the frequencies.

The method of drawing histogram is given below

**Example:**

Draw a Histogram for the following table which represent the age of spectators in a Cricket Match.

Age in years | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 |

No. of Spectators | 4 | 6 | 12 | 10 | 8 | 2 |

**Solution:**

The class intervals are equal in length. Let us denote the age of spectators along x axis and the number of spectators along y axis. The Histogram is given below.

Drawing Histogram when class intervals are not continuous

**Example:**

Yield per acre | 11 – 15 |
16 – 20 |
21 -25 |
26 -30 |
31 – 35 |
36 – 40 |

No. of rice fields |
3 |
5 |
18 |
15 |
6 |
4 |

**Solution:**

We observe that the class intervals are not continuous. If we draw histogram for this data we will get gaps between the class intervals. But in Histogram class intervals need to be continuous and there should not be gap between the intervals so we make the class intervals continuous by inserting an adjustment factor.

Adjustment Factor= ½(Lower limit of the class interval- Upper limit of the class interval)

=1/2 (16-15)

= 0.5

In the above class interval we subtract 0.5 from each lower limit and add 0.5 to each upper limit. Thus we rewrite the given data into the following data

Yield per acre |
10.5 – 15.5 |
15.5 – 20.5 |
20.5 -25.5 |
25.5 – 30.5 |
30.5 – 35.5 |
35.5 – 40.5 |

No. of rice fields |
3 |
5 |
18 |
15 |
6 |
4 |

**Frequency Polygon:**

Frequency Polygon is another method of representing the data graphically.

Draw a Histogram for the given continuous data. Mark the middle points at the top of adjacent rectangle. If we join these middle points successively by line segment we obtain polygon. This polygon is called “Frequency Polygon”.

**Example:**

Draw a frequency polygon for the below data

Marks |
0-10 |
10-20 |
20-30 |
30-40 |
40-50 |
50-60 |
60-70 |
70-80 |
80-90 |
90-100 |

No. of students |
5 |
4 |
6 |
8 |
5 |
7 |
4 |
9 |
5 |
7 |

**Solution:**

Mark the marks of students along x axis and the number of students along y axis. Draw a Histogram for the above data. Now mark the midpoint at the top of each adjacent rectangles. The midpoints are joined with the help of a ruler.

Note that the first and last edges of the Frequency Polygon meet at the vertical edges of the first and the last rectangles.

Very informative, thank you for the post