Question

Just before a referendum on a school​ budget, a local newspaper polls 373 voters to predict whether the budget will pass. Suppose the budget has the support of 53% of the voters. What is the probability that the​ newspaper's sample will lead it to predict​ defeat?

Answer 1

Answer on Question #67660 – Math – Statistics and Probability

Question

Just before a referendum on a school budget, a local newspaper polls 373 voters to predict whether the budget will pass. Suppose the budget has the support of 53% of the voters. What is the probability that the newspaper's sample will lead it to predict defeat?

Solution

$$\begin{array}{l} \mu_p = 0.53; \\ \sigma = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.53 \times 0.47}{373}} = 0.0258. \\ P(\hat{p} > 0.5) = P\left(Z > \frac{X - \mu_p}{\sigma}\right) = P\left(Z > \frac{0.5 - 0.53}{0.0258}\right) = P(Z > -1.16) = \\ = 1 - P(Z < -1.16) = 0.8775. \end{array}$$

Answer: 0.8775.

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