Use the empirical rule (68-95-99.7) to determine if the following maximum daily rainfall totals could be considered to be normally distributed prompt.

1.1, 1.2, 1.5, 1.55, 1.71, 1.75, 1.9, 1.94, 1.99, 2.0, 2.05, 2.08, 2.12, 2.18, 2.2, 2.25, 2.36, 2.5, 2.61, 2.81, 2.88, 2.95, 3.0, 3.03, 3.07, 3.16, 3.25, 3.62. (note: they are already in order)

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brennarfa

23.04.2022

Mathematics

$2\sin 2\theta - 3 \cos2\theta - 3\sin\theta + 3 = \sin\theta(4\cos\theta + 6\sin\theta - 3)$ left-hand side $\begin{aligned} \text{lhs} & = 2\sin 2\theta - 3 \cos2\theta - 3\sin\theta + 3 \\ & = 2\boxed{\sin 2\theta} - 3 \boxed{\cos2\theta} - 3\sin\theta + 3 \end{aligned}$ use identities: $\begin{aligned} \sin 2\theta & = 2\sin\theta\cos\theta \\ \cos 2\theta & = 1 - 2\sin^2\theta \end{aligned}$ (using that form of cosine double angle because there's one 1 cosine on the right-hand side. if we used another form, there'd be too many cosines) $\begin{aligned} \text{lhs} & = 2(2\sin\theta\cos\theta)- 3(1 - 2\sin^2\theta) - 3\sin\theta + 3 \\ & = 4\sin\theta\cos\theta - 3 + 6\sin^2\theta - 3\sin\theta + 3 \\ & =4\sin\theta\cos\theta - \boxed{3} + 6\sin^2\theta - 3\sin\theta + \boxed{3} \end{aligned}$ those +3 and -3 cancel each other out (add to become 0) $\begin{aligned} \text{lhs} & =4\sin\theta\cos\theta + 6\sin^2\theta - 3\sin\theta \end{aligned}$ factor out $\sin\theta$ out of everything $\begin{aligned} \text{lhs} & =4\sin\theta\cos\theta + 6\sin^2\theta - 3\sin\theta \\ & = \sin\theta\left(4\cos\theta + 6\sin\theta - 3\right) \\ & = \text{rhs} \end{aligned}$

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