Questions: Suppose 48% of the banks in Switzerland are private organizations. If a sample of 484 banks is selected, what is the probability that the sample proportion of private banks will differ from the population proportion by greater than 3%? Round your answer to four decimal places.

Solution Steps

The standard error (SE) of the sample proportion is calculated using the formula:

\[ SE = \sqrt{\frac{p(1 - p)}{n}} \]

where \( p = 0.48 \) (the population proportion) and \( n = 484 \) (the sample size). Substituting the values, we find:

\[ SE = \sqrt{\frac{0.48(1 - 0.48)}{484}} \approx 0.0227 \]

The range for the sample proportion is defined by the margin of error of \( 3\% \) (or \( 0.03 \)). Thus, the lower and upper bounds are:

\[ \text{Range Start} = p - 0.03 = 0.48 - 0.03 = 0.45 \] \[ \text{Range End} = p + 0.03 = 0.48 + 0.03 = 0.51 \]

To find the probability that the sample proportion falls within the range \( [0.45, 0.51] \), we calculate the Z-scores for the range limits:

\[ Z_{start} = \frac{0.45 - 0.48}{0.0227} \approx -1.3203 \] \[ Z_{end} = \frac{0.51 - 0.48}{0.0227} \approx 1.3203 \]

Using the standard normal distribution, we find:

\[ P(Z_{start} < Z < Z_{end}) = \Phi(Z_{end}) - \Phi(Z_{start}) \approx 1.0 - 0.0 = 1.0 \]

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

\[ P(\text{outside range}) = 1 - P(\text{within range}) = 1 - 1.0 = 0.0 \]

Final Answer