Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 205.6-cm and a standard deviation of 1.1-cm. For shipment, 10 steel rods are bundled together. Find P25, which is the average length separating the smallest 25% bundles from the largest 75% bundles. P25

Answer 1

P25 = 205.365cm

Explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 205.6, \sigma = 1.1, n = 10, s = (1.1)/(√(10)) = 0.34785`

Find P25, which is the average length separating the smallest 25% bundles from the largest 75% bundles. P25

This is the value of X when Z has a pvalue of 0.25. So X when Z = -0.675.

So

`Z = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`-0.675 = (X - 205.6)/(0.34785)`

`X - 205.6 = -0.675*0.34785`

`X = 205.365`

P25 = 205.365cm