Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 143.6-cm and a standard deviation of 0.8-cm. For shipment, 41 steel rods are bundled together. Round all answers to four decimal places if necessary.

1. 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 143.4-cm and 143.5-cm.
• For a bundled of 41 rods, find the probability that the average length is between 143.4-cm and 143.5-cm.
• For part d), is the assumption of normal necessary? NoYes

Answer 1

Solution:

Given: The lengths of the steel rods are normally distributed with a mean of 143.6-cm and a standard deviation of 0.8-cm.

Part a) What is the distribution of X?

The distribution of X is Normal( mean = , standard deviation = )

That is:

X ~ N( 143.6 , 0.8 )

Part b) What is the distribution of ?

The distribution of is Normal with mean of is and standard deviation of   is:

Thus

~ N( 143.6 , 0.1249)

Part c) For a single randomly selected steel rod, find the probability that the length is between 143.4-cm and 143.5-cm.

P( 143.4 < X < 143.5 ) = ............?

Find z score for x = 143.4

Thus

P( 143.4 < X < 143.5 ) = P( -0.25 < Z < -0.13 )

P( 143.4 < X < 143.5 ) = P( Z < -0.13 ) - P( Z < -0.25 )

Thus Look in z table for z = -0.1 and 0.03 as well as for z = -0.2 and 0.05 and find corresponding area.

P( Z< -0.13) =0.4483

P( Z<-0.25) = 0.4013

thus

P( 143.4 < X < 143.5 ) = P( Z < -0.13 ) - P( Z < -0.25 )

P( 143.4 < X < 143.5 ) = 0.4483 - 0.4013

P( 143.4 < X < 143.5 ) = 0.0470

Part d) For a bundled of 41 rods, find the probability that the average length is between 143.4-cm and 143.5-cm.

Find z score :

thus we get:

Look in z table for z = -0.8 and 0.00as well as for  z = -1.6 and 0.00  and find corresponding area.

P( Z < -0.80 0 = 0.2119

and

P( Z< -1.60) = 0.0548

thus

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

Since X is already Normally distributed, so it is not necessary to assumption of Normal distribution

thus answer is: No.