In a frequency distribution, class mark of a class is 25 and the lower boundary of the class is 9. Find the class length.

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Solution

The correct option is C 32
Let,
lower boundary = x = 9
upper boundary = y
$C l a s s m a r k s = \frac{upper boundary + lower boundary}{2}$
25 = $\frac{x+y}{2}$

y + 9 = 50
y = 50 - 9 = 41
Class length = upper boundary - lower boundary = y - x = 41 - 9 = 32.

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In a frequency distribution, class mark of a class is 25 and the lower boundary of the class is 9. Find the class length.

The correct option is C 32Let, lower boundary = x = 9 upper boundary = y Class marks=upper boundary + lower boundary 2 25 = x+y2 y + 9 = 50 y = 50 - 9 = 41 Class length = upper boundary - lower boundary = y - x = 41 - 9 = 32.

The correct option is C 32
Let,
lower boundary = x = 9
upper boundary = y
$C l a s s m a r k s = \frac{upper boundary + lower boundary}{2}$
25 = $\frac{x+y}{2}$

y + 9 = 50
y = 50 - 9 = 41
Class length = upper boundary - lower boundary = y - x = 41 - 9 = 32.