Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 209.5-cm and a standard deviation of 1.1-cm. For shipment, 44 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 209.4-cm and 209.6-cm. For a bundled of 44 rods, find the probability that the average length is between 209.4-cm and 209.6-cm. For part d), is the assumption of normal necessary? NoYes

Answer 1

Answer:

a)

Given,

X~N(u , /sqrt(n))

X~N(209.5 , 1.1/sqrt(1))

X~N(209.5 , 1.1)

b)

P(209.4 < X < 209.6) = P((209.4 - 209.5)/1.1 < (x-u)/s < (209.6 - 209.5)/1.1)

= P(-0.09 < z < 0.09)

= P(z < 0.09) - P(z < -0.09)

= 0.5358563 - 0.4641436 [since from z table]

= 0.0717

c)

Given,

X~N(u , /sqrt(n))

X~N(209.5 , 1.1/sqrt(44))

X~N(209.5 , 0.1658)

P(209.4 < X < 209.6) = P((209.4 - 209.5)/0.1658 < (x-u)/s < (209.6 - 209.5)/0.1658)

= P(-0.60 < z < 0.60)

= P(z < 0.60) - P(z < -0.60)

= 0.7257469 - 0.2742531 [since from z table]

= 0.4515

d)

Yes, the normal distribution is necessary.