Questions: Suppose 48% of the population has a retirement account. If a random sample of size 632 is selected, what is the probability that the proportion of persons with a retirement account will differ from the population proportion by greater than 3%? Round your answer to four decimal places.

Solution Steps

The population proportion of individuals with a retirement account is given as \( p = 0.48 \). The population standard deviation for a proportion is calculated using the formula:

\[ \sigma = \sqrt{p(1 - p)} = \sqrt{0.48 \times (1 - 0.48)} = \sqrt{0.48 \times 0.52} \approx 0.4996 \]

The sampling standard deviation (standard error) for the sample size \( n = 632 \) is calculated as follows:

\[ \sigma_{\text{sampling}} = \frac{\sigma}{\sqrt{n}} = \frac{0.4996}{\sqrt{632}} \approx 0.0199 \]

We want to find the probability that the sample proportion differs from the population proportion by more than \( 3\% \) (or \( 0.03 \)). This means we are interested in the range:

\[ \left( p - 0.03, p + 0.03 \right) = \left( 0.48 - 0.03, 0.48 + 0.03 \right) = \left( 0.45, 0.51 \right) \]

Using the Z-scores for the bounds:

\[ Z_{\text{start}} = \frac{0.45 - 0.48}{0.0199} \approx -1.505 \] \[ Z_{\text{end}} = \frac{0.51 - 0.48}{0.0199} \approx 1.505 \]

The cumulative probabilities for these Z-scores are:

\[ P = \Phi(Z_{\text{end}}) - \Phi(Z_{\text{start}}) = \Phi(1.505) - \Phi(-1.505) \approx 1.0 \]

The probability that the sample proportion differs from the population proportion by more than \( 3\% \) is given by:

\[ P_{\text{outside}} = 1 - P_{\text{within}} = 1 - 1.0 = 0.0000 \]

Final Answer