We are given that the cashier’s order processing time is normally distributed with a mean of 276 seconds and a standard deviation of 38 seconds. We need to find the proportion of orders that are processed in less than 240 seconds.

Step 1: Calculate the Z-score

The Z-score formula is given by:

$Z = \frac{X - \mu}{\sigma}$

Where:

• $X = 240$ (the value we're analyzing),
• $\mu = 276$ (mean),
• $\sigma = 38$ (standard deviation).

Substituting the values:

$Z = \frac{240 - 276}{38} = \frac{-36}{38} \approx -0.947$

Step 2: Look up the Z-score in a standard normal distribution table

For a Z-score of approximately -0.95, the cumulative probability (which gives us the proportion of orders processed in less than 240 seconds) is approximately 0.17.

Conclusion:

The proportion of orders processed in less than 240 seconds is closest to 0.17, which corresponds to option A.

Would you like any further details or have any questions?

Here are 5 related questions you might be interested in:

1. How do Z-scores help in interpreting normal distributions?
2. What is the empirical rule for standard normal distributions?
3. How can the cumulative distribution function (CDF) be used in this context?
4. What is the significance of negative Z-scores?
5. How would the result change if the standard deviation was smaller?

Tip: Remember, Z-scores can help standardize values and compare data points from different normal distributions.