Q75 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics

The variance of the data : $1,2,3,5,8,13,17$ is approximately

  • Correct answer
    A
    $31.14$
  • B
    $29.57$
  • C
    $30.62$
  • D
    $32.71$

Solution

$\begin{aligned} & \text { } \frac{\Sigma x_i}{n}=\frac{49}{7}=7 \\ & \Sigma x_i^2=1+4+9+25+64+169+289=561 \\ & \sigma^2=\frac{\Sigma x_i^2}{n}-\left(\frac{\Sigma x_i}{n}\right)^2=\frac{561}{7}-49=\frac{218}{7}=31.14\end{aligned}$

Hence, the correct answer is option: 'A'.

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