We are given that the a particular candidate for public office is in fact favored by p = 48% of all registered voters. A polling organization is about to take a simple random sample of voters and will use the sample proportion to estimate p.

Suppose that the polling organization takes a simple random sample of 500 voters.

Let $\hat p$ = sample proportion

The z-score probability distribution for sample proportion is given by;

Z = $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion

p = population proportion = 48%

n = sample of voters = 500

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

So, probability that the sample proportion will be greater than 0.5 is given by = P( $\hat p$ > 0.50)

P( $\hat p$ > 0.50) = P( $\frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < $\frac{0.50-0.48}{\sqrt{\frac{0.50(1-0.50)}{500} } }$ ) = P(Z < 0.89) = 0.8133

Now, in the z table the P(Z $\leq$ x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.89 in the z table which has an area of 0.8133.

Therefore, probability that the sample proportion will be greater than 0.50 is 0.8133.

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4 years ago