SOLUTION: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 257.2-cm and a standard deviation of 1.5-cm. For shipment, 12 steel rods are bu

Question 1206525: A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 257.2-cm and a standard deviation of 1.5-cm. For shipment, 12 steel rods are bundled together.
Find the probability that the average length of a randomly selected bundle of steel rods is between 257.7-cm and 258.1-cm.
P(257.7-cm < M < 258.1-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 by Theo(13342) (Show Source): You can put this solution on YOUR website!
mean is 257.2 cm.
standard deviation is 1.5 cm
sample size is 12.

standard error = population standard deviation / sqrt(sample size) = 1.5 / sqrt(12) = .433013.

you want to know the probability that the mean of the sample will be between 257.7 and 258.1 cm.

it appears the problem is looking for z-score results, so we'll go with that.

z-score formula is z = (x - m) / s

z is the z-score
x is the sample mean
m is the population mean
s is the standard error

standard error = standard deviation / sqrt(sample size) = 1.5 / sqrt(12) = .433013 rounded to 6 decimal places.

you want to find the z-scores for sample means of 257.7 and 258.1.

z-score formula for sample mean of 257.7 is z = (257.7 - 257.2) / .433013 = 1.1547.

z-score formula for sample mean of 258.1 is z = (258.1 - 257.2) / .433013 = 2.0785.

you can use a z-score calculator, such as the one at https://davidmlane.com/hyperstat/z_table.html

use of that calculator tells you that the area between those 2 z-score is equal to .1053.

that's the probability that a sample of size 12 will have a mean length of between 257.7 and 258.1 centimeters.

here's what that looks like.

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