Question

The manufacturer of a new compact car claims the miles per gallon (mpg) for the fuel economy is normally distributed with mean = 25.9 and a standard deviation = 9.5. What is the probability that one new compact car selected at random has a fuel economy of at least 23?

a) 0.7257
b) 0.7881
c) 0.2743
d) 0.8413

Answer 1

To find the probability of a new compact car having a fuel economy of at least 23 mpg, we calculate the z-score for 23 and use the standard normal distribution to find the probability. However, the calculated probability does not match any of the given choices.

Step-by-step explanation:

The student is asking about the probability that a randomly selected new compact car will have a fuel economy of at least 23 miles per gallon (mpg) given a normal distribution with a mean of 25.9 mpg and a standard deviation of 9.5. To find this probability, we need to calculate the z-score for 23 mpg and then use the standard normal distribution to find the probability of a z-score being greater than the calculated z-score.

First, we calculate the z-score for 23 mpg:

Z = (X - μ) / σ
Z = (23 - 25.9) / 9.5
Z = -2.9 / 9.5
Z = -0.3053

Now, we look up this z-score in a standard normal distribution table or use a calculator with normal distribution functions to find the probability that Z is greater than -0.3053.

The probability associated with a z-score of -0.3053 is approximately 0.62. Since we want the probability of obtaining a value of at least 23, we subtract this from 1 to get the remaining area under the curve:

P(X ≥ 23) = 1 - P(Z < -0.3053)
P(X ≥ 23) = 1 - 0.62
P(X ≥ 23) = 0.38

Since none of the choices given (a through d) match the calculated probability, there might be a mistake in the calculation or in the provided choices. The correct approach, however, has been demonstrated.