Statistics quetion

Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 160.4-cm and a standard deviation of 2.4-cm. For shipment, 25 steel rods are bundled together.

Find P₁₂, which is the average length separating the smallest 12% bundles from the largest 88% bundles.

With steps please

Answer 1

z score = (mean - mu)/(SD/sqrt(25) = (mean - 160.4)/0.48

find z-score corresponding to 12th percentile: -1.175

substitute -1.175 into first equation and solve for mean:

-1.175 = (mean - 160.4)/0.48

mean = (-1.175 * 0.48) + 160.4 = 159.836

Answer 2

First, we have to assume that the lengths of the rods within a bundle are uncorrelated (that is, if a longer-than-average rod is part of a bundle, that tells me nothing about the lengths of the other 24 rods in the bundle). Second, I will assume that the question is asking about the total length of rods in a bundle, rather than the length of the longest rod in the bundle (the latter question is much trickier, and I can update my answer if that is what was intended).

Note that we are looking for the width of an interval but not where the interval is located, so the mean length is irrelevant to the question; only the variance matters. When we add 25 identical, uncorrelated normal distributions together, the variance is 25 times the variance of a single normal distribution. (Note that this does not mean that the standard deviation is multiplied by 25.)

The single normal distribution has a standard deviation of 2.4cm, meaning its variance is 2.4² cm² = 5.76cm². So the variance of the 25 rods is 25*5.76cm² = 144cm². The standard deviation of the sum is the square root of the variance, or 12cm.

The way the problem is written, it sounds like we need to find the average of the values in the bottom 12% of a normal distribution and the top 88% of a normal distribution. This is quite complicated, but if that's the question, you can find an answer of how to calculate it here: https://math.stackexchange.com/questions/1546085/mean-of-a-portion-of-a-normal-distribution.

The question I think (and hope) you're asking is the difference between the 12th and 88th percentiles. Here we can just plug a mean of 0 and standard deviation of 12 into a normal distribution calculator (https://homepage.divms.uiowa.edu/~mbognar/applets/normal.html) to find that the 12th percentile is at -14.1 and the 88th percentile is at 14.1, so the total average length separating them is 28.2cm.