​In a frequency distribution, if the mid-value of the class is 35 and the value of the lower boundary is 30, then the value of its upper boundary is:

Correct option is C

Given:

Mid-value of the class = 35

Lower boundary of the class = 30

Concept Used:

The mid-value (class midpoint) of a class interval is calculated as:

​$$\text{Mid-value} = \frac{\text{Lower boundary} + \text{Upper boundary}}{2}$$​​

Solution:

​$$35 = \frac{30 + \text{Upper boundary}}{2}$$​​

​$$35 \times 2 = 30 + \text{Upper boundary}$$​​

​$$70 = 30 + \text{Upper boundary}$$​​

Upper boundary=70−30=40

Option(c) is right.