KCET · Maths · Functions
The standard deviation of the data \(6,7,8,9\), 10 is
- A \(\sqrt{2}\)
- B \(\sqrt{10}\)
- C 2
- D 10
Answer & Solution
Correct Answer
(A) \(\sqrt{2}\)
Step-by-step Solution
Detailed explanation
Given data \(6,7,8,9,10\)
\(\begin{aligned}
\bar{x} &=\frac{6+7+8+9+10}{5}=8 \\
\mathrm{SD} &=\sqrt{\frac{\sum x_{i}^{2}}{n}-(\bar{x})^{2}} \\
\mathrm{SD} &=\sqrt{\frac{6^{2}+7^{2}+8^{2}+9^{2}+10^{2}}{5}-(8)^{2}} \\
\mathrm{SD} &=\sqrt{\frac{36+49+64+81+100-320}{5}} \\
\mathrm{SD} &=\sqrt{\frac{330-320}{5}}=\sqrt{\frac{10}{5}}=\sqrt{2}
\end{aligned}\)
\(\begin{aligned}
\bar{x} &=\frac{6+7+8+9+10}{5}=8 \\
\mathrm{SD} &=\sqrt{\frac{\sum x_{i}^{2}}{n}-(\bar{x})^{2}} \\
\mathrm{SD} &=\sqrt{\frac{6^{2}+7^{2}+8^{2}+9^{2}+10^{2}}{5}-(8)^{2}} \\
\mathrm{SD} &=\sqrt{\frac{36+49+64+81+100-320}{5}} \\
\mathrm{SD} &=\sqrt{\frac{330-320}{5}}=\sqrt{\frac{10}{5}}=\sqrt{2}
\end{aligned}\)
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