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Q.
In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is
Statistics
Solution:
Let $x$ be the upper limit and $y$ be the lower limit.
Since the mid value of the class is 10
$\therefore \, \frac{x+y}{2} = 10 \:\: \Rightarrow \:\:\:\:\:\:\: x + y = 20 $ ....(1)
and $x - y = 6$ (width of the class = 6) ...(2)
By solving (1) and (2), we get y = 7.
Hence, lower limit of the class is 7.