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JEE Main 20249 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
9 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2024 · 9 April · Evening Shift · Mathematics · Statistics

If the variance of the frequency distribution array |c|c|c|c|c|c|c|...

If the variance of the frequency distribution xc2c3c4c5c6cf211111\begin{array}{|c|c|c|c|c|c|c|} \hline x & c & 2 c & 3 c & 4 c & 5 c & 6 c \\ \hline f & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array}is 160, then the valueofc∈Nvalue ofc \in N is$

Options

  1. A7
  2. B8
  3. C5
  4. D6

Correct answer

A. 7

Step-by-step solution

xC2C3C4C5C6Cf211111\begin{array}{|c|c|c|c|c|c|c|} \hline \mathrm{x} & \mathrm{C} & 2 \mathrm{C} & 3 \mathrm{C} & 4 \mathrm{C} & 5 \mathrm{C} & 6 \mathrm{C} \\ \hline \mathrm{f} & 2 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array} xˉ=(2+2+3+4+5+6)C7=22C7\bar{x}=\frac{(2+2+3+4+5+6) C}{7}=\frac{22 C}{7} Var⁡(x)=c2(2+22+32+42+52+62)7−(22c7)2=92c27−c2×48449=(644−484)c249=160c249160=160×c249⇒c=7\begin{aligned} & \operatorname{Var}(x)=\frac{\mathrm{c}^2\left(2+2^2+3^2+4^2+5^2+6^2\right)}{7} \\ & -\left(\frac{22 \mathrm{c}}{7}\right)^2 \\ & =\frac{92 \mathrm{c}^2}{7}-\mathrm{c}^2 \times \frac{484}{49} \\ & =\frac{(644-484) \mathrm{c}^2}{49}=\frac{160 \mathrm{c}^2}{49} \\ & 160=\frac{160 \times \mathrm{c}^2}{49} \Rightarrow \mathrm{c}=7\end{aligned}
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