Question \#3 (12 points)

Weights of a certain model of fully loaded gravel trucks follow a normal distribution with mean $\mu=8.6$ tons and standard deviation $\sigma=0.5$ tons. What is the probability that a fully loaded truck of this model is:
a.) More than 7.5 tons?
b.) Less than 9.7 tons?
c.) Between 8.5 and 8.7 tons?

Step 1 ：Given that the weights of a certain model of fully loaded gravel trucks follow a normal distribution with mean \( \mu = 8.6 \) tons and standard deviation \( \sigma = 0.5 \) tons.

Step 2 ：For part a, we are asked to find the probability that a fully loaded truck of this model is more than 7.5 tons. We first calculate the z-score, which is \( z_a = \frac{X_a - \mu}{\sigma} = \frac{7.5 - 8.6}{0.5} = -2.2 \). The probability is then given by \( 1 - \) the cumulative distribution function at \( z_a \), which is approximately 0.986.

Step 3 ：For part b, we are asked to find the probability that a fully loaded truck of this model is less than 9.7 tons. We first calculate the z-score, which is \( z_b = \frac{X_b - \mu}{\sigma} = \frac{9.7 - 8.6}{0.5} = 2.2 \). The probability is then given by the cumulative distribution function at \( z_b \), which is approximately 0.986.

Step 4 ：For part c, we are asked to find the probability that a fully loaded truck of this model is between 8.5 and 8.7 tons. We first calculate the z-scores, which are \( z_{c1} = \frac{X_{c1} - \mu}{\sigma} = \frac{8.5 - 8.6}{0.5} = -0.2 \) and \( z_{c2} = \frac{X_{c2} - \mu}{\sigma} = \frac{8.7 - 8.6}{0.5} = 0.2 \). The probability is then given by the cumulative distribution function at \( z_{c2} \) minus the cumulative distribution function at \( z_{c1} \), which is approximately 0.159.

Step 5 ：Final Answer: The probability that a fully loaded truck of this model is more than 7.5 tons is approximately \(\boxed{0.986}\), less than 9.7 tons is approximately \(\boxed{0.986}\), and between 8.5 and 8.7 tons is approximately \(\boxed{0.159}\).