The gas prices in 2019 were normally distributed with a mean of \( \$ 2.99 \) and a standard deviation of \( \$ 0.19 \) Consider a randomly selected day in 2019 . You may find it helpful to use the Normal Distribution tool. Enter all answers as a percent rounded to two decimal places. Include the \( \% \) symbol in your answer. (a) What is the probability of finding gas prices above \( \$ 3.69 \) ? \( P(X>\$ 3.69)= \) (b) What is the probability of finding gas prices between \( \$ 2.92 \) and \( \$ 3.22 \) ? \( P(\$ 2.92

Ask by Rodriquez Floyd. in the United States

Mar 14,2025

To calculate the probabilities in each scenario, we can standardize the values using the z-score formula: \[ z = \frac{(X - \mu)}{\sigma} \] where \( \mu \) is the mean and \( \sigma \) is the standard deviation. (a) For \( P(X > \$ 3.69) \): 1. Calculate the z-score: \[ z = \frac{(3.69 - 2.99)}{0.19} \approx 3.684 \] 2. Using a z-table or standard normal distribution calculator, we find: \[ P(Z > 3.684) \approx 0.0001 \] So, \( P(X > \$ 3.69) \approx 0.01\% \). (b) For \( P(\$ 2.92 < X < \$ 3.22) \): 1. Calculate the z-scores: \[ z_1 = \frac{(2.92 - 2.99)}{0.19} \approx -0.368 \] \[ z_2 = \frac{(3.22 - 2.99)}{0.19} \approx 1.211 \] 2. Using the z-table: - For \( z_1 \): \( P(Z < -0.368) \approx 0.3557 \) - For \( z_2 \): \( P(Z < 1.211) \approx 0.8871 \) 3. Therefore, \[ P(2.92 < X < 3.22) = P(Z < 1.211) - P(Z < -0.368) = 0.8871 - 0.3557 \approx 0.5314 \] This results in \( P(2.92 < X < 3.22) \approx 53.14\% \). (c) For \( P(X < 2.87) \): 1. Calculate the z-score: \[ z = \frac{(2.87 - 2.99)}{0.19} \approx -0.632 \] 2. Using the z-table: \[ P(Z < -0.632) \approx 0.2643 \] Therefore, \( P(X < 2.87) \approx 26.43\% \). To summarize: (a) \( P(X > \$ 3.69) \approx 0.01\% \) (b) \( P(\$ 2.92 < X < \$ 3.22) \approx 53.14\% \) (c) \( P(X < \$ 2.87) \approx 26.43\% \)