Class Boundaries are the data values that separate classes. These are not part of the classes or the data set. The class boundary is the middle point of the upper-class limit of one class and the lower class limit of the subsequent class. Each class has an upper and a lower class boundary. Find the definition of the class boundaries or actual class limits and example questions in the below-mentioned sections.

## What is Class Boundary?

If we have different classes of data, then it has an upper-class limit and lower class limit which means the smaller and larger values. A class boundary is the midpoint of the upper-class limit of one class and the lower class limit of the subsequent class.

Lower class boundary = (lower class limit of the concerned class + upper-class limit of the previous class)/2

Upper class boundary = (upper-class limit of the concerned class + lower class limit of the subsequent class)/2

For the nonoverlapping class intervals,

The actual lower limit = lower limit – ½ x gap

The actual upper limit = upper limit + ½ x gap

### Class Limit

The class limit can be defined as the minimum and maximum values contain in a class interval.

The minimum value is called the lower class limit and the maximum value is called the upper-class limit.

### Solved Examples on True Class Limits

**Example 1:**

Class | Frequency |
---|---|

0 – 9 | 2 |

10 – 19 | 5 |

20 – 29 | 7 |

Find the lower, upper class boundaries?

**Solution:**

The class boundary of 0 – 9 is

Upper class boundary = (upper-class limit of the concerned class + lower class limit of the subsequent class)/2

= \(\frac { (9 + 10) }{ 2 } \)

= \(\frac { 19 }{ 2 } \)

Lower class boundary = (lower class limit of the concerned class + upper class limit of the previous class)/2

= \(\frac { (0 + 0) }{ 2 } \)

= 0

The class boundary of 10 – 19 is

Upper class boundary = \(\frac { (19 + 20) }{ 2 } \)

= \(\frac { 39 }{ 2 } \)

Lower Class boundary = \(\frac { (10 + 9) }{ 2 } \)

= \(\frac { 19 }{ 2 } \)

**Example 2:**

If the class marks of two consecutive overlapping intervals of equal size in distribution are 94 and 104 then find the corresponding intervals.

**Solution:**

The difference between 104 and 94 = 104 – 94 = 10

Therefore, the class intervals are (94 – \(\frac { 10 }{ 2 } \)) – (94 + \(\frac { 10 }{ 2 } \)) and (104 – \(\frac { 10 }{ 2 } \)) – (104 + \(\frac { 10 }{ 2 } \))

= (94 – 5) – (94 + 5) and (104 – 5) and (104 + 5)

= 89 – 99 and 99 – 109.

**Example 3:**

Weight in Kg (Class Interval) | Frequency |
---|---|

44-48 | 3 |

49-53 | 4 |

54-58 | 5 |

59-63 | 7 |

64-68 | 9 |

69-73 | 8 |

Find the upper, lower class boundaries?

**Solution:**

Lower Class Boundary of the first class interval = 44 – \(\frac { (49 – 48) }{ 2 } \)

= 44 – \(\frac { 1 }{ 2 } \)

= 44 – 0.5

= 43.5

Upper Class Boundary = 48 + \(\frac { (49 – 48) }{ 2 } \)

= 48 + \(\frac { 1 }{ 2 } \)

= 48 + 0.5

= 48.5.

### FAQs on Class Boundaries

**1. What is a Class Boundary?**

A class boundary is the midpoint of the upper-class limit of one class and the lower class limit of the subsequent class.

**2. What is the difference between class limits and class boundaries?**

In the case of the class limit, the upper extreme value of the first class interval and the lower extreme value of the next class interval are not equal. However, in the class boundary, the upper extreme value of the first class interval and the lower extreme value of the next class interval are equal.

**3. How to find Class Boundaries?**

Follow the below-listed steps to calculate the Class Boundaries easily. They are along the lines

- Subtract the upper class limit for the first class from the lower class limit for the second class.
- Divide the result by two.
- Subtract the result from the lower class limit and add the result to the upper class limit for each class.