To solve this problem, we use the properties of a normal distribution.

Problem Summary

We have:

• A normally distributed length of items with:
  • Mean ($\mu$) = 16.7 inches
  • Standard deviation ($\sigma$) = 4.8 inches
• We want to find the probability that an item's length is less than 16.6 inches.

Solution

To calculate this probability, we first find the z-score for 16.6 inches. The z-score formula is:

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

where:

• $X = 16.6$ (the value we're interested in),
• $\mu = 16.7$,
• $\sigma = 4.8$.

Plugging in the values:

$z = \frac{16.6 - 16.7}{4.8} = \frac{-0.1}{4.8} \approx -0.0208$

Now, we use the standard normal distribution table or a calculator to find the probability corresponding to $z = -0.0208$.

Using the Standard Normal Table

For $z = -0.0208$:

• The corresponding probability is approximately 0.4917.

Conclusion

The probability that a randomly chosen item is less than 16.6 inches long is approximately 0.4917 or 49.17%.

Would you like further details or have any questions?

Here are five related questions:

1. What is the probability of selecting an item longer than 16.7 inches?
2. How would the probability change if the standard deviation were smaller?
3. How do we interpret the z-score in terms of standard deviations?
4. How do you find probabilities for ranges in normally distributed data?
5. What is the probability of selecting an item between 15 and 20 inches long?

Tip: In a normal distribution, probabilities can be quickly found using z-scores and standard normal tables or software, simplifying complex probability calculations.