Consider the following frequency distribution.

Question

Consider the following frequency distribution. If mean $=\dfrac{309}{22}$ and median $=14$, then the value $(a-b)^2$ is equal to ..........

Answer: 4

Accepted Answer

Mean: $\dfrac{3a+9b+504}{a+b+26}=\dfrac{309}{22} \implies 81a+37b=1018$

Median: $12+\dfrac{\frac{a+b+26}{2}-(a+b)}{12}	imes6=14 \implies$ simplifying gives $a+b=18$

Solving $81a+37b=1018$ with $a+b=18$: $a=8, b=10$

$(a-b)^2 = (8-10)^2=4$

Solutions · 1