A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of $173.9 \mathrm{~cm}$ and a standard deviation of $2.3 \mathrm{~cm}$. For shipment, 23 steel rods are bundled together.

Answer the following rounding to three decimals where appropriate.
Find the probability that the average length of a randomly selected bundle of steel rods is less than $173.76 \mathrm{~cm}$.
\[
P(M< 173.76)=
\]
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Step 1 ：Calculate the Z-score using the formula: \(Z = \frac{X - \mu}{\sigma/\sqrt{n}}\)

Step 2 ：Substitute the given values into the formula: \(Z = \frac{173.76 - 173.9}{2.3/\sqrt{23}}\)

Step 3 ：Simplify the equation to get: \(Z = \frac{-0.14}{2.3/\sqrt{23}}\)

Step 4 ：Further simplify the equation to get: \(Z = \frac{-0.14}{0.479}\)

Step 5 ：Solve the equation to get: \(Z = -0.292\)

Step 6 ：Find the probability that the Z-score is less than -0.292 using a standard normal distribution table or calculator

Step 7 ：The probability is approximately 0.385

Step 8 ：So, the probability that the average length of a randomly selected bundle of steel rods is less than 173.76 cm is approximately 0.385, or 38.5%

Step 9 ：\(\boxed{0.385}\)