please help me solve this

Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 177.9-cm and a standard deviation of 1.1-cm. For shipment, 18 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is greater than 178.4-cm.

P(M > 178.4-cm) = 

Enter your answer as a number accurate to 4 decimal places. Answers obtained using exact z-scores or z-scores rounded to 3 decimal places are accepted.

Answer 1

Find the sample standard deviation based on the given sample size and the population standard deviation:

s = σ/√n = 1.1/√18 ≈ 0.2593

Now find the z-score using the sample standard deviation to standardize the normal distribution:

z = (M-x̄)/s = (178.4-177.9)/(1.1/√18) ≈ 1.928

Hence, P(M>178.4) = P(Z>1.928) ≈ 1 - 0.9756 = 0.0244

This means that the probability that the average length of a randomly selected bundle of steel rods is greater than 178.4 cm would be 0.0244, or about 1 in every 40 rods. Hope this helped!