Let $\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{N}$ and $\mathrm{a}< \mathrm{b}< \mathrm{c}$. Let the mean, the mean deviation about the mean and the variance of the 5 observations $9,25, a, b, c$ be 18, 4 and $\frac{136}{5}$, respectively. Then $2 a+b-c$ is equal to ________
Answer (integer)
33
Solution
<p>$$\begin{aligned}
& a, b, c \in N \\
& a< b < c \\
& \text { Mean }=18 \\
& \frac{9+25+a+b+c}{5}=18 \\
& 34+a+b+c=90 \\
& a+b+c=56
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{|9-18|+|25-18|+|a-18|+|b-18|+|c-18|}{5}=4 \\
& 9+7+|a-18|+|b-18|+|c-18|=20 \\
& |a-18|+|b-18|+|c-18|=4 \\
& \frac{136}{5}=\frac{706+a^2+b^2+c^2}{5}-(18)^2 \\
& \Rightarrow 136=706+a^2+b^2+c^2-1620 \\
& \Rightarrow a^2+b^2+c^2=1050 \\
& \text { Consider } a<19 < b< c \\
& \text { Solving } a=17, b=19, c=20 \\
& 2 a+b-c \\
& 34+19-20 \\
& =33
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Statistics · Topic: Measures of Central Tendency
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