Find the value of t if the standard deviation of 2t, 3t, 4t, 5t, and 6t is √2
\(mean(x) = \frac{20t}{5}\)
= S.D = \(\sqrt{\frac{((\overline{x} - x)^2}{\text{n}}}\)
\(\sqrt{2}\) = \(\sqrt{\frac{10t^2}{5}}\)
2 = \(\frac{10t^2}{5}\)
10t\(^2\) = 10
t\(^2\) = 1
t = \(\sqrt{1}\) = 1
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