Find the value of t if the standard deviation of 2t, 3t, 4t, 5t, and 6t is √2
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Correct Answer: Option A
Explanation:
\(mean(x) = \frac{20t}{5} = 4t\\standard\hspace{1mm}deviation=\sqrt{\frac{10t^2}{5}}\\\sqrt{2}=\sqrt{\frac{10t^2}{5}}\\\sqrt{2}=\sqrt{2t^2}\\2 = 2t^2\\t^2=1\\t=1\)
\(mean(x) = \frac{20t}{5} = 4t\\standard\hspace{1mm}deviation=\sqrt{\frac{10t^2}{5}}\\\sqrt{2}=\sqrt{\frac{10t^2}{5}}\\\sqrt{2}=\sqrt{2t^2}\\2 = 2t^2\\t^2=1\\t=1\)