Question

Suppose that \(65 \%\) of all registered voters in a certain area favor a 7 -day waiting period before purchase of a handgun. Among 225 randomly selected voters, what is the probability that a. At least 150 favor such a waiting period? b. More than 150 favor such a waiting period? c. Fewer than 125 favor such a waiting period?

Short answer

The probabilities that a. At least 150 voters favor such a waiting period is approximately 0.2999, b. More than 150 voters favor such a waiting period is approximately 0.2468, c. Fewer than 125 voters favor such a waiting period is approximately 0.0015. These are approximations using normal distribution and actual values may differ slightly.

Step-by-step solution

Normal Approximation Criteria

Verify if binsize * probability * (1 - probability) is greater than 5. If true, we can apply the normal approximation. For our case: \(225 * 0.65 * 0.35 = 51.1875\), so we can apply the normal approximation.

Calculate Mean and Standard Deviation

Calculate mean (\(mu\)) and standard deviation (\(sigma\)) of the distribution using formulas: \(mu = n*p\) and \(sigma = sqrt(n*p*(1 - p)\). where n is the number of trials (voters here - 225), p is the probability (0.65 in this case). Mean, \(mu = 225 * 0.65 = 146.25\) and Standard deviation, \(sigma = sqrt(225 * 0.65 * 0.35) = 7.153\).

Convert to Z-Scores

Convert the given numbers (150, 151, and 125) into z-scores using formula: \(z = (X - mu) / sigma\). z-score for 150 voters: \(z = (150 - 146.25) / 7.153 = 0.526\), for 151 voters: \(z = (151 - 146.25) / 7.153 = 0.664\) and for 125 voters: \(z = (125 - 146.25) / 7.153 = -2.97\).

Find Probabilities Using Z-Table

a. For 'at least 150 voters'', we are looking for \(P(X >= 150)\) or \(1 - P(z < 0.526)\) which equals to \(1 - 0.7001 = 0.2999\). b. For 'more than 150 voters', we are interested in \(P(X > 150)\) which equals to \(P(z > 0.664)\) and equals to \(1 - P(z < 0.664) = 1 - 0.7532 = 0.2468\). c. For 'fewer than 125 voters', we need to find \(P(X < 125)\) or \(P(z < -2.97)\) equals to 0.0015.