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JEE Main 20248 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2024 · 8 April · Evening Shift · Mathematics · Statistics

Let a, b, c ∈ N and a < b < c . Let the mean, the mean deviation...

Let a,b,c∈N\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathrm{N} and a<b<c\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,c9,25, \mathrm{a}, \mathrm{b}, \mathrm{c} be 18,4 and 1365\frac{136}{5}, respectively. Then 2a+b−c2 \mathrm{a}+\mathrm{b}-\mathrm{c} is equal to__________

Correct answer

0

Step-by-step solution

a,b,c∈Na<b<ca, b, c \in N \quad a < b < c x‾=\overline{\mathrm{x}}= mean =9+25+a+b+c5=18=\frac{9+25+\mathrm{a}+\mathrm{b}+\mathrm{c}}{5}=18 a+b+c=56a+b+c=56 Mean deviation =Σ∣xi−x‾∣n=4=\frac{\Sigma\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|}{\mathrm{n}}=4 =9+7+∣18−a∣+∣18−b∣+∣18−c∣=20=∣18−a∣+∣18−b∣+∣18−c∣=4\begin{aligned} & =9+7+|18-\mathrm{a}|+|18-\mathrm{b}|+|18-\mathrm{c}|=20 \\ & =|18-\mathrm{a}|+|18-\mathrm{b}|+|18-\mathrm{c}|=4\end{aligned} Variance =Σ∣xi−x‾∣2n=1365=\frac{\Sigma\left|\mathrm{x}_{\mathrm{i}}-\overline{\mathrm{x}}\right|^2}{\mathrm{n}}=\frac{136}{5} =81+49+∣18−a∣2+∣18−b∣2+∣18−c∣2=136=(18−a)2+(18−b)2+(18−c)2=6\begin{aligned} & =81+49+|18-a|^2+|18-b|^2+|18-c|^2=136 \\ & =(18-a)^2+(18-b)^2+(18-c)^2=6\end{aligned} Possible values (18−a)2=1,(18−b)2=1(18−c)2=4(18-\mathrm{a})^2=1, \quad(18-\mathrm{b})^2=1 \quad(18-\mathrm{c})^2=4 a<b<c\mathrm{a} < \mathrm{b} < \mathrm{c} so \quad18−a=118−b=−118−c=−2a=17 b=19c=20\begin{array}{ccc}18-a=1 & 18-b=-1 & 18-c=-2 \\ \mathrm{a}=17 & \mathrm{~b}=19 & \mathrm{c}=20\end{array}$a+b+c=562 \mathrm{a}+\mathrm{b}-\mathrm{c} 34=19-20=33$
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