A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 232-cm and a standard deviation of 1.2-cm. For shipment, 26 steel rods are bundled together.

Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 232-cm and a standard deviation of 1.2-cm. For shipment, 26 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is less than 231.9-cm.

P(M < 231.9-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

Using formulas for the standard error and for z statistic we calculate the Z stat and compare to the normal distribution.

population mean (µ) = 232 , std.dev. of bundle (σ) = 1.2. size of bundle = 26

std.dev. of mean (std.error) = 1.2/√26 = .235339

Formula for Z-score = x-µ / se. so z for P(M < 231.9-cm). = (231.9 - 232) / .235339. = -0.1/.235339.

= -.424919

Using Z value calculator (https://www.calculator.net/z-score-calculator.html)

P(M < 231.9-cm) =  0.336