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In order to solve the following inequality for $n$, where $0 < \theta < 1$, they use a trick that I do not understand

$n \geq (\frac{1.96}{0.03})^2 \theta(1 - \theta)$

Since $\theta$ is unknown we take $\theta=0.5$ so the inequality is true for all $0 < \theta < 1$. Thus

$n \geq (\frac{1.96}{0.03})^2 (0.5)^2 = 1067.1$

Why are they allowed to set $\theta = 0.5$? Why does that value of $\theta$ make the inequality hold for all $0 < \theta <1$?

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  • $\begingroup$ In general, drawing a little picture is a good idea. In this case, plotting $x \mapsto x(1-x)$ will give a hint. $\endgroup$ – copper.hat Jun 9 at 20:56
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The value of $\theta(1-\theta)$ is maximal when we set $\theta=\frac12$. That is the reasoning justifying that step.

(Think of a sketch of the parabola $x(1-x)$).

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