Question

Just before a referendum on a school​ budget, a local newspaper polls 354 voters to predict whether the budget will pass. Suppose the budget has the support of 55​% of the voters. What is the probability that the​ newspaper's sample will lead it to predict​ defeat?

Answer 1

Answer:

3.07% probability that the​ newspaper's sample will lead it to predict​ defeat

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For a proportion p with n voters, we use $\mu = p, \sigma = \sqrt{\frac{p(1-p)}{n}}}$

In this problem, we have that:

$p = 0.55, n = 354$

So

$\mu = 0.55, \sigma = \sqrt{\frac{0.55*0.45}{345}}} = 0.0268$

What is the probability that the​ newspaper's sample will lead it to predict​ defeat?

That is, a percentage of 50% or less, which is the pvalue of Z when X = 0.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{0.5 - 0.55}{0.0268}$

$Z = -1.87$

$Z = -1.87$ has a pvalue of 0.0307.

3.07% probability that the​ newspaper's sample will lead it to predict​ defeat