The problem presented in the image is about calculating the percentage of truck load times that are either less than 223 minutes or more than 259 minutes, given a normal distribution model with a mean of 247 minutes and a standard deviation of 10 minutes.

Step-by-step solution:

1. Define the variables:

   • Mean $\mu = 247$ minutes
   • Standard deviation $\sigma = 10$ minutes
   • We are tasked with finding the percentages for:
     • Less than 223 minutes
     • More than 259 minutes

2. Convert to Z-scores: The Z-score formula is: $Z = \frac{X - \mu}{\sigma}$ where $X$ is the value you're comparing against.

   • For 223 minutes: $Z_1 = \frac{223 - 247}{10} = \frac{-24}{10} = -2.4$
   • For 259 minutes: $Z_2 = \frac{259 - 247}{10} = \frac{12}{10} = 1.2$

3. Find the cumulative probability: Using standard normal distribution tables or a calculator:

   • $P(Z < -2.4)$: The cumulative probability for $Z = -2.4$ is approximately 0.0082, meaning 0.82% of the values are less than 223 minutes.
   • $P(Z > 1.2)$: The cumulative probability for $Z = 1.2$ is 0.8849. To find the percentage greater than 259 minutes, we calculate: $P(Z > 1.2) = 1 - 0.8849 = 0.1151$ This means 11.51% of the values are more than 259 minutes.

4. Final answer: Adding the percentages together, we get: $0.82\% + 11.51\% = 12.33\%$ So, approximately 12.33% of the load times are either less than 223 minutes or more than 259 minutes.

Would you like more details or have any questions?

Here are 5 related questions:

1. How do you calculate Z-scores for other time intervals?
2. What is the probability of loading times between 223 and 259 minutes?
3. How does changing the mean or standard deviation affect the probabilities?
4. Can you explain how normal distribution applies to real-world problems?
5. What tools can be used to quickly calculate probabilities from Z-scores?

Tip: Always double-check the Z-score calculations and use a reliable standard normal distribution table or calculator for accuracy.