The fuel efficiency rating for a mid-size car is normally distributed with a mean of 32 and a standard deviation of 0.8. What is the probability that the rating for a selected mid-size car would be:
-More than 33.2?
A)0.4332
B)0.0668
C)0.9332
D)0.8664
E)0.1336

Question

Quizplus · Accepted Answer

To find the probability that the rating is more than 33.2, we first calculate the Z-score: Z = (X - μ) / σ = (33.2 - 32) / 0.8 = 1.5. Using a standard normal distribution table, the probability of Z being less than 1.5 is approximately 0.9332. Therefore, the probability of being more than 33.2 is 1 - 0.9332 = 0.0668.

The Fuel Efficiency Rating for a Mid-Size Car Is Normally