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A fair random poll of 900 voters finds 51% for candidate P and 49% for candidate Q…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

A fair random poll of 900 voters finds 51% for candidate P and 49% for candidate Q. A newspaper says that P is ahead. Of the 900 voters, 225 are under 25. The newspaper also says that young voters back Q, by 54% to 46%. Use the quick rule to find the margin for the whole poll and the margin for the young voters, each to one decimal place. Is either claim safe? Use the fact that a lead has a margin of about twice the poll's margin.

Need a reminder? The key ideaShow

The margin of luck of a fair random poll is about $\dfrac{100}{\sqrt n}$ percentage points, for about 95 polls in 100. For a margin $m$ , take $n = \left(\dfrac{100}{m}\right)^2$ . A group of $n_g$ inside the poll has the margin $\dfrac{100}{\sqrt{n_g}}$ .

The worked solution

Have you tried it?

Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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