To solve this problem using the Central Limit Theorem (CLT), we need to follow these steps:

Step 1: Identify the key values

• Population mean ($\mu$) = 1.73 days
• Population standard deviation ($\sigma$) = 1.73 days
• Sample size ($n$) = 108
• Sample mean ($\bar{x}$) = 1.53 days (the value we are comparing to)

Step 2: Calculate the standard error of the mean (SE)

The standard error (SE) is the standard deviation of the sample mean distribution. It is calculated as:

$SE = \frac{\sigma}{\sqrt{n}}$

Substituting the given values:

$SE = \frac{1.73}{\sqrt{108}} \approx \frac{1.73}{10.3923} \approx 0.1665$

Step 3: Find the z-score

The z-score represents how many standard errors the sample mean is from the population mean. It is calculated using the formula:

$z = \frac{\bar{x} - \mu}{SE}$

Substituting the values:

$z = \frac{1.53 - 1.73}{0.1665} \approx \frac{-0.20}{0.1665} \approx -1.2012$

Step 4: Find the probability

To find the probability corresponding to a z-score of $-1.2012$, we look up the z-score in the standard normal distribution table (or use a calculator). The probability of a z-score less than $-1.2012$ is approximately 0.1156.

Thus, the probability that the mean shipping time for the 108 shipments is less than 1.53 days is approximately 0.1156.

Final Answer:

$P(\bar{x} < 1.53) \approx 0.1156$

Would you like more details or further clarification?

Relative Questions:

1. What is the Central Limit Theorem and how does it apply to sample means?
2. How do you interpret the standard error of the mean in practical terms?
3. What is the significance of the z-score in a normal distribution?
4. How would the probability change if the sample size were smaller?
5. What is the relationship between sample size and standard error?

Tip:

The Central Limit Theorem allows us to approximate the distribution of the sample mean as normal, even if the original population distribution is not normal, provided the sample size is large enough (usually $n \geq 30$).