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

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

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 = (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{(p(1-p))/(n)}}`

In this problem, we have that:

`p = 0.55, n = 354`

So

`\mu = 0.55, \sigma = \sqrt{(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 = (X - \mu)/(\sigma)`

`Z = (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