Answer to Question #324387 in Statistics and Probability for HunterJ

Question

Just before a referendum on a school​ budget, a local newspaper polls 405 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

We are given;

population proportion; μ = 55% = 0.55

Sample size;n = 405

The conditions are;

10% conditon: sample size is less than 10% of the population size

Success or failure condition; np = 405 x 0.55 = 223 and n(1 - p) = 405(1 - 0.55) = 182

Both values are greater than 10

Randomization condition; we assume that the voters were randomly selected.

So the conditions are met.

Now, the standard deviation is gotten from;

$\sigma=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.55(1-0.55)}{405}}=0.025$

where;

p is the population proportion

n is the sample size

σ is standard deviation

Now to find the z-value, we'll use;

P(X>0.5)=P(Z>(0.5-0.55)/0.025)=P(Z>-2)=1-P(Z<-2)=1-0.02275=0.97725

Thus, the probability the newspaper’s sample will lead them to predict defeat is 0.97725