A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 139 cm and a standard deviation of 2 cm. Find , which is the length separating the shortest 90% rods from the longest 10%. Round your answer to at least one decimal place.

1. Identify the mean \mu and standard deviation \sigma :
\mu = 139 \, \text{cm}
\sigma = 2 \, \text{cm}

2. Look for the z-score corresponding to the 90th percentile in the standard normal distribution table, denoted as z_{0.90} :
z_{0.90} \approx 1.28

3. Use the z-score formula to find L :
L = \mu + z_{0.90} \times \sigma
L = 139 + 1.28 \times 2

4. Calculate the result:
L = 139 + 2.56 = 141.56 \, \text{cm}

Therefore, the length separating the shortest 90% rods from the longest 10% is:
L = 141.56 \, \text{cm}