Central Limit Theorem 8.0

Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 212.8-cm and a standard deviation of 0.7-cm. For shipment, 24 steel rods are bundled together.

Find the probability that the average length of a randomly selected bundle of steel rods is less than 213.2-cm.

P(M < 213.2-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

xbar = average length of steel rod; population mean 212.8 cm and population standard deviation 0.7 cm. Where distribution is expressed as N(212.8, 0.7), where N is Normal.

want to find P(M < 213.2 cm)

Probability is found using standardized normal probability table, which is based on N(0,1) distribution, whose input is standardized Z variable.

Standardized Z variable = (M - pop mean)/(std dev/sqrt(sample size))

Compute standardized z variable corresponding to M = 213.2

Z = (213.2 - 212.8)/(0.7/sqrt(24)) = 2.797

So :P(M < 213.2) = P(Z < 2.797) = 0.9974