The problem asks to find the probability that an item is less than 12.6 inches long when the item's length is normally distributed with a mean ($\mu$) of 13 inches and a standard deviation ($\sigma$) of 1.3 inches. We can use the standard normal distribution to find this probability.

Step-by-step solution:

1. Define the given values:

   • Mean ($\mu$) = 13 inches
   • Standard deviation ($\sigma$) = 1.3 inches
   • We need to find $P(X < 12.6)$, where $X$ is the random variable representing the length of the item.

2. Standardize the variable: To use the standard normal distribution, we first convert the value 12.6 into a z-score, which is the number of standard deviations away from the mean. The formula for the z-score is:

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

   Substituting the values:

   $z = \frac{12.6 - 13}{1.3} = \frac{-0.4}{1.3} \approx -0.31$

3. Find the probability: Using a standard normal distribution table or a calculator, we look up the z-score of $-0.31$. The corresponding probability is approximately 0.3783.

Thus, the probability that the item's length is less than 12.6 inches is approximately 0.3783, or 37.83%.

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Do you want more details on how to interpret the z-table or have any questions?

Here are some related questions:

1. How do we calculate probabilities for values above a certain length?
2. What is the probability for an item being more than 14 inches long?
3. How does changing the standard deviation affect the distribution?
4. How do we find the probability between two lengths, e.g., between 12 and 14 inches?
5. What is the cumulative distribution function (CDF) and how does it relate to normal distributions?

Tip: When working with normal distributions, it's helpful to always standardize the data (convert to z-scores) to use standard normal distribution tables or calculators effectively.