Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 269.4-cm and a standard deviation of 1.4-cm. For shipment, 37 steel rods are bundled together. Round all answers to four decimal places if necessary.

What is the distribution of X? X ~ N

What is the distribution of ¯x? ¯x ~ N

For a single randomly selected steel rod, find the probability that the length is between 269.3-cm and 269.4-cm.

For a bundled of 37 rods, find the probability that the average length is between 269.3-cm and 269.4-cm.

For part d), is the assumption of normal necessary? yes or no

Answer 1

X ~ N(269.4, 1.4)
xbar ~ N(269.4, 1.4/sqrt(37)) = N(269.4, 0.2302)

c)
z = (x - μ)/σ
z1 = (269.3 - 269.4)/1.4 = -0.07
z2 = (269.4 - 269.4)/1.4 = 0

Therefore, we get
P(269.3 <= X <= 269.4) = P((269.4 - 269.4)/1.4) <= z <= (269.4 - 269.4)/1.4)
= P(-0.07 <= z <= 0) = P(z <= 0) - P(z <= -0.07)
= 0.5 - 0.4721
= 0.0279

d)
z = (x - μ)/σ
z1 = (269.3 - 269.4)/0.2302 = -0.43
z2 = (269.4 - 269.4)/0.2302 = 0

Therefore, we get
P(269.3 <= X <= 269.4) = P((269.4 - 269.4)/0.2302) <= z <= (269.4 - 269.4)/0.2302)
= P(-0.43 <= z <= 0) = P(z <= 0) - P(z <= -0.43)
= 0.5 - 0.3336
= 0.1664

e)
Yes