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  • Consider the following frequency distribution : <img alt="\begin{array}{lccccc}\text { Class: } & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 \\ \text { Frequency : } & \text { a } & \text { b } & 12 & 9 & 5\end{array

Consider the following frequency distribution :
\begin{array}{lccccc}\text { Class: } & 0-6 & 6-12 & 12-18 & 18-24 & 24-30 \\ \text { Frequency : } & \text { a } & \text { b } & 12 & 9 & 5\end{array}
If mean =\frac{309}{22} and median =14, then the value (a-b)^{2} is equal to_______.
 

Answers (1)

best_answer
Class \text{fi} \text{xi} \text{fixi}
0-6 a 3 3a
6-12 b 9 9b
12-18 12 15 180
18-24 9 21 189
24-30 5 27 135
  \text{N=26+a+b}   \sum \text{fixi}=504+3 a+9 b

\begin{aligned} \text { Mean } &=\frac{504+3 a+9 b}{26+a+b}=\frac{309}{22} \\ & \Rightarrow 81 a+37 b=1018 ...........\quad \text { (i) } \end{aligned}

\begin{gathered} \text { Median }=12+\frac{\frac{a+b+26}{2}-(a+b)}{12} \cdot 6=14 \\ \Rightarrow a+b=18........\text { (ii) } \end{gathered}

Solving we get a=8, b=10

\Rightarrow \quad(a-b)^{2}=(8-10)^{2}=4

Hence, the answer is (4).

Posted by

Kuldeep Maurya

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