KCET · Maths · Statistics
The standard deviation of the numbers 31 , \(32,33 \ldots 46,47\) is
- A \(\sqrt{\frac{17}{12}}\)
- B \(\sqrt{\frac{47^{2}-1}{12}}\)
- C \(2 \sqrt{6}\)
- D \(4 \sqrt{3}\)
Answer & Solution
Correct Answer
(C) \(2 \sqrt{6}\)
Step-by-step Solution
Detailed explanation
Given, numbers : \(31,32,33, \ldots, 46,47\). The standard deviation of \(31,32,33, \ldots, 47\) will be the same as those of \(1,2,3, \ldots, 17\). (decreasing each item by 30 )
\(\because \quad \mathrm{SD}=\sqrt{\frac{n^{2}-1}{12}}\)
Here, \(\quad n=17\)
\(\begin{aligned}
\therefore \quad \mathrm{SD} &=\sqrt{\frac{17^{2}-1}{12}}=\sqrt{\frac{288}{12}} \\
&=\sqrt{24}=2 \sqrt{6}
\end{aligned}\)
\(\because \quad \mathrm{SD}=\sqrt{\frac{n^{2}-1}{12}}\)
Here, \(\quad n=17\)
\(\begin{aligned}
\therefore \quad \mathrm{SD} &=\sqrt{\frac{17^{2}-1}{12}}=\sqrt{\frac{288}{12}} \\
&=\sqrt{24}=2 \sqrt{6}
\end{aligned}\)
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