Final answer:
To find the probability of the sample mean exceeding 30.75 metric tons, calculate the z-score and consult the standard normal distribution. Option B) 0.1587 is the correct answer.
Step-by-step explanation:
The probability that the sample mean impact force from a sample of 4 cars (x) exceeds 30.75 metric tons can be found using the standard normal distribution. Since the impact forces are normally distributed with a mean (μ) of 30 metric tons and a standard deviation (σ) of 1.5 metric tons, we would first find the standard error of the mean by dividing the standard deviation by the square root of the sample size (n=4). The standard error (SE) would be σ/sqrt(n) = 1.5 / 2 = 0.75 metric tons.
Next, we calculate the z-score for 30.75 metric tons which is (30.75 - 30) / 0.75 = 1. The probability of x being greater than 30.75 is the area to the right of z=1 in the standard normal distribution. Consulting the z-table or using statistical software, we find this area corresponds to the option B) 0.1587.