The lengths of nails are normally distributed, with a mean ($\mu$) of 4.81 cm and a standard deviation ($\sigma$) of 0.05 cm. To find the two lengths that separate the top 10% and the bottom 10%, we are essentially looking for the 10th and 90th percentiles of this normal distribution.

1. Determine the z-scores for the 10th and 90th percentiles. These values correspond to the points in the normal distribution where 10% of the data falls below (bottom 10%) and 10% of the data falls above (top 10%).

From standard normal distribution tables or using statistical software, we know:

• The z-score for the 10th percentile is approximately $-1.28$.
• The z-score for the 90th percentile is approximately $+1.28$.

2. Convert the z-scores into actual lengths using the formula for a normally distributed variable:

$x = \mu + z \cdot \sigma$

• For the bottom 10%: $x_{10} = 4.81 + (-1.28) \cdot 0.05 = 4.81 - 0.064 = 4.75 \, \text{cm}$

• For the top 10%: $x_{90} = 4.81 + (1.28) \cdot 0.05 = 4.81 + 0.064 = 4.87 \, \text{cm}$

Thus, the two lengths that separate the top 10% and the bottom 10% are approximately:

• Bottom 10%: 4.75 cm
• Top 10%: 4.87 cm

These lengths could be used as limits to identify nails that should be rejected.

Would you like more details on this or have any questions?

Here are 5 related questions:

1. How do you calculate z-scores for other percentiles?
2. What happens to the rejection limits if the standard deviation increases?
3. How would you use a normal distribution table to find specific z-scores?
4. Can you explain the impact of changing the mean on the rejection limits?
5. How can you use this concept in quality control for other manufacturing processes?

Tip: Understanding z-scores helps in quickly determining probabilities and percentiles for normally distributed data.