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. \( 0.25 \% \) percentage, include the \% symbol. (b) Find the probability that \( 63 \% \) to \( 64 \% \) of the 38 households don't own a car. I Round proportions to four decimal places or percentages to two decimal places. If entering a percentage, include the \% symbol.

To solve these probabilities, we can model the household ownership with a binomial distribution, where: - \( n = 38 \) (the number of households surveyed) - \( p = 0.55 \) (the probability that a household does not own a car) We can first calculate the mean (\( \mu \)) and standard deviation (\( \sigma \)) of the distribution: \[ \mu = np = 38 \times 0.55 = 20.9 \] \[ \sigma = \sqrt{np(1-p)} = \sqrt{38 \times 0.55 \times 0.45} \approx \sqrt{9.405} \approx 3.07 \] (a) To find the probability that 30 or more households do not own a car, we need to convert this into a z-score: \[ z = \frac{x - \mu}{\sigma} = \frac{30 - 20.9}{3.07} \approx 2.96 \] Now, we look up the z-score of 2.96 in the standard normal distribution table, which corresponds to approximately \( 0.9985 \). This means: \[ P(X \geq 30) = 1 - P(Z < 2.96) \approx 1 - 0.9985 = 0.0015 \] Thus, the probability that 30 or more households do not own a car is approximately \( 0.15 \% \). (b) For the probability that \( 63\% \) to \( 64\% \) of the 38 households do not own a car, we convert these percentages to counts: - \( 63\% \) of 38 households is \( 0.63 \times 38 \approx 23.94 \) (round this to 24) - \( 64\% \) of 38 households is \( 0.64 \times 38 \approx 24.32 \) (round this to 24) Now, we already calculated the probabilities for 24 households. We will find the z-scores: For \( x = 24 \): \[ z = \frac{24 - 20.9}{3.07} \approx 1.01 \] Using the z-table, \( P(Z < 1.01) \approx 0.8438 \). For \( x = 25 \): \[ z = \frac{25 - 20.9}{3.07} \approx 1.32 \] Using the z-table, \( P(Z < 1.32) \approx 0.9066 \). Thus, the probability that \( 63\% \) to \( 64\% \) of households do not own a car is approximately: \[ P(24 \leq X < 25) = P(Z < 1.32) - P(Z < 1.01) \approx 0.9066 - 0.8438 = 0.0628 \] Converting this to a percentage, we find: Approximately \( 6.28\% \).