Question

A company produces steel rods. The lengths of the steel rods are normally distributed with a mean of 111.4-cm and a standard deviation of 0.5-cm. For shipment, 23 steel rods are bundled together. Find the probability that the average length of a randomly selected bundle of steel rods is between 111.2-cm and 111.4-cm. P(111.2-cm < ¯ x < 111.4-cm) =

Answer 1

Answer:

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 111.4, \sigma = 0.5, n = 23, s = \frac{0.5}{\sqrt{23}} = 0.1043$

Find the probability that the average length of a randomly selected bundle of steel rods is between 111.2-cm and 111.4-cm.

This is the pvalue of Z when X = 111.4 subtracted by the pvalue of Z when X = 111.2. So

X = 111.4

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{111.4 - 111.4}{0.1043}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5.

X = 111.2

$Z = \frac{X - \mu}{s}$

$Z = \frac{111.2 - 111.4}{0.1043}$

$Z = -1.92$

$Z = -1.92$ has a pvalue of 0.0274.

0.5 - 0.0274 = 0.4726

P(111.2-cm < ¯ x < 111.4-cm) = 0.4726