Q75 / 160 · AP EAMCET 2019 (23 Apr Shift 1) · Mathematics

The variance of the following data is \(\begin{array}{c|c|c|c|c|c|c|c|c|c|c} \hline \boldsymbol{x}_{\boldsymbol{i}} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \boldsymbol{f}_{\boldsymbol{i}} & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 \\ \hline \end{array}\)

  • A
    10
  • B
    9
  • C
    8
  • Correct answer
    D
    6

Solution

According to given data, \(\begin{array}{ccccc} \hline \boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}} & \boldsymbol{x}_{\boldsymbol{i}} \boldsymbol{f}_{\boldsymbol{i}} & \overline{\boldsymbol{x}}-\boldsymbol{x}_{\boldsymbol{i}} & \boldsymbol{f}_{\boldsymbol{i}}\left(\overline{\boldsymbol{x}}-\boldsymbol{x}_{\boldsymbol{i}}\right)^{\mathbf{2}} \\ \hline 1 & 1 & 1 & 6 & 36 \\ \hline 2 & 2 & 4 & 5 & 50 \\ \hline 3 & 3 & 9 & 4 & 48 \\ \hline 4 & 4 & 16 & 3 & 36 \\ \hline 5 & 5 & 25 & 2 & 20 \\ \hline 6 & 6 & 36 & 1 & 6 \\ \hline 7 & 7 & 49 & 0 & 0 \\ \hline 8 & 8 & 64 & 1 & 8 \\ \hline 9 & 9 & 81 & 2 & 36 \\ \hline 10 & 10 & 100 & 3 & 90 \\ \hline \end{array}\) \(\begin{aligned} \because \quad N & =\Sigma f_i=55, \bar{x}=\frac{\Sigma x_i f_i}{N} \\ & =\frac{10 \times 11 \times 21}{6 \times 55}=7 \\ \therefore \quad \sigma^2 & =\frac{\Sigma f_i\left(\bar{x}-x_i\right)^2}{N} \\ & =\frac{36+50+48+36+20+6+8+36+90}{55} \\ & =\frac{330}{55}=6 \end{aligned}\) Hence, option (d) is correct.

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

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