Question:
The mid-value of a class interval is 42 and the class size is 10. The lower and upper limits are
(a) 37−47
(b) 37.5−47.5
(c) 36.5−47.5
(d) 36.5−46.5
Solution:
(a) 37–47
Let the lower limit be x.
Here,
Class size = 10
∴ Upper limit = Class size + Lower limit
Upper limit = (x + 10)
Mid value of the class interval = 42
$\therefore \frac{x+x+10}{2}=42$
$\Rightarrow \frac{2 x+10}{2}=42$
$\Rightarrow 2 x+10=84$
$\Rightarrow 2 x=74$
$\Rightarrow x=37$
Thus, we have :
Lower limit $=37$
Upper limit $=37+10=47$