Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 226.9-cm and a standard deviation of 1.6-cm. For shipment, 47 steel rods are bundled together. Round all answers to four decimal places if necessary.

What is the distribution of XX? XX ~ N(,)

What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)

For a single randomly selected steel rod, find the probability that the length is between 227-cm and 227.1-cm.

For a bundled of 47 rods, find the probability that the average length is between 227-cm and 227.1-cm.

For part d), is the assumption of normal necessary? NoYes

Answer 1

Given that, mean (μ) = 226.9 cm and

standard deviation = 1.6 cm

a) Therefore, the distribution of X is, X ~ N(226.9, 1.6)

b) sample size (n) = 47

The mean and standard deviation of the sampling distribution of the sample means are,

Therefore, the distribution of X is,

c) We want to find, P(227 < X < 227.1)

Therefore, required probability is 0.0278

Note : if we used TI-83 plus calculator then we will get probability = 0.0248

d) We want to find,

Therefore, required probability is 0.1387

Note : if we used TI-83 plus calculator then we will get probability = 0.1384

e) Yes, for part d) the assumption of normal is necessary.