Across the five boroughs, it's estimated that \( 55 \% \) of households in New York City do not own a car. You may use the Normal Distribution tool. Mean should not be rounded, standard deviation may be rounded to four decimal places. (a) Find the probability that 30 or more households in a randomly selected survey of 38 households don't own a car. percentage, include the \( \% \) symbol.

To solve this problem, we first need to define parameters for our normal distribution. The proportion of households that do not own a car is \( p = 0.55 \). For a sample size \( n = 38 \), we can calculate the mean (expected number of households that don't own a car) and the standard deviation. **Mean (\( \mu \)):** \[ \mu = n \cdot p = 38 \cdot 0.55 = 20.9 \] **Standard Deviation (\( \sigma \)) of the binomial distribution:** \[ \sigma = \sqrt{n \cdot p \cdot (1 - p)} = \sqrt{38 \cdot 0.55 \cdot 0.45} \approx 3.0264 \] Using the normal approximation, we need to find the probability of observing 30 or more households not owning a car. We can use the continuity correction factor since we are approximating a binomial distribution with a normal distribution. So we are looking for: \[ P(X \geq 30) = P(X > 29.5) \] Applying the Z-score formula: \[ Z = \frac{X - \mu}{\sigma} = \frac{29.5 - 20.9}{3.0264} \approx 2.8342 \] Now we can find the probability using the Z-table or a calculator: \[ P(Z > 2.8342) \] The corresponding area to the left of \( Z = 2.8342 \) is approximately \( 0.9974 \). Hence, \[ P(Z > 2.8342) = 1 - 0.9974 = 0.0026 \] Converting this probability into a percentage: \[ 0.0026 \times 100 \approx 0.26\% \] The probability that 30 or more households in a randomly selected survey of 38 households don't own a car is approximately \( 0.26\% \).