Automobile repair costs continue to rise with the average cost now at \( \$ 367 \) per repair. + Assume that the cost for an automobile repair is normally distributed with a standard deviation of \( \$ 88 \). Answer the following questions about the cost of automobile repairs. (a) What is the probability that the cost will be more than \( \$ 470 \) ? (Round your answer to four decimal places.) (b) What is the probability that the cost will be less than \( \$ 220 \) ? (Round your answer to four decimal places.) (c) What is the probability that the cost will be between \( \$ 220 \) and \( \$ 470 \) ? (Round your answer to four decimal places.) (d) If the cost for your car repair is in the lower \( 5 \% \) of automobile repair charges, what is your maximum possible cost in dollars? (Round your answer to the nearest cent.) \( \$ 222.24 \)

To find the answers to the questions about automobile repair costs given the normal distribution, we first need to standardize the values using the z-score formula, which is: \[ z = \frac{X - \mu}{\sigma} \] where \( X \) is the value in question, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. **(a)** For finding the probability that the cost will be more than \( \$470 \): 1. Calculate the z-score for \( 470 \): \[ z = \frac{470 - 367}{88} \approx 1.171 \] 2. Looking up \( z = 1.171 \) in the standard normal distribution table, we find the area to the left, which is approximately \( 0.8808 \). 3. Therefore, the probability that the cost is more than \( \$470 \) is: \[ P(X > 470) = 1 - P(Z < 1.171) \approx 1 - 0.8808 = 0.1192 \] Rounded to four decimal places: **0.1192**. **(b)** For the probability that the cost will be less than \( \$220 \): 1. Calculate the z-score for \( 220 \): \[ z = \frac{220 - 367}{88} \approx -1.670 \] 2. Looking up \( z = -1.670 \) in the standard normal distribution table, we find the area to the left, which is approximately \( 0.0478 \). 3. Therefore, the probability that the cost is less than \( \$220 \) is: \[ P(X < 220) \approx 0.0478 \] Rounded to four decimal places: **0.0478**. **(c)** For the probability that the cost will be between \( \$220 \) and \( \$470 \): 1. We already found \( P(X < 220) \approx 0.0478 \) and \( P(X < 470) \approx 0.8808 \) (from part a). 2. Therefore, \[ P(220 < X < 470) = P(X < 470) - P(X < 220) = 0.8808 - 0.0478 = 0.8330 \] Rounded to four decimal places: **0.8330**. **(d)** To find the maximum possible cost in the lower \( 5\% \) of automobile repair charges: 1. We need the z-score that corresponds to \( 5\% \), which is approximately \( -1.645 \). 2. Now, we rearrange the z-score formula to find \( X \): \[ X = \mu + (z \cdot \sigma) = 367 + (-1.645 \cdot 88) \approx 367 - 144.56 \approx 222.44 \] Rounded to the nearest cent: **\$222.44**. Thus the answers are: - (a) 0.1192 - (b) 0.0478 - (c) 0.8330 - (d) \$222.44