(i) Find the class mark of the class 90 – 120.

Question:

(i) Find the class mark of the class 90 – 120.
(ii) In a frequency distribution, the mid-value of the class is 10 and width of the class is 6. Find the lower limit of the class.
(iii) The width of each of five continuous classes in a frequency distribution is 5 and lower class limit of the lowest class is 10. What is the upper class limit of the highest class?
(iv) The class marks of a frequency distribution are 15, 20, 25, ... . Find the class corresponding to the class mark 20.
(v) In the class intervals 10 – 20, 20 – 30, find the class in which 20 is included.

Solution:

(i) class mark $=\frac{\text { upper limit+lower limit }}{2}=\frac{120+90}{2}=\frac{210}{2}=105$

(ii) mid-value = 10
width = 6
Let the lower limit of the class be x
upper limit = x + 6

class $\mathrm{mark} / \mathrm{mid}-\mathrm{value}=\frac{\text { upper limit+lower limit }}{2}$

$\frac{x+(x+6)}{2}=10$

$\Rightarrow x=7$

(iii) width = 5
lower class limit of lowest class = 10
The classes will be 10-15, 15-20, 20-25, 25-30, 30-35.
Upper class limit of the highest class = 35.
(iv) Class marks = 15, 20, 25, ...
class size = 20 -">- 15 = 5
Let lower limit of class be x.

$\frac{x+(x+5)}{2}=20$

$\Rightarrow x=17.5$

Thus, the class is 17.5-22.5. 
(v) 20 will be included in the class interval 20-30. 

 

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