Question

A manufacturer knows that their items have a normally distributed length, with a mean of 13.1 inches, and standard deviation of 4.1 inches. If 25 items are chosen at random, what is the probability that their mean length is less than 11.1 inches

Answer 1

Answer:

0.73% probability that their mean length is less than 11.1 inches

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean $\mu$ and standard deviation $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation, which is also called standard error $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 13.1, \sigma = 4.1, n = 25, s = \frac{4.1}{\sqrt{25}} = 0.82$

What is the probability that their mean length is less than 11.1 inches

This is the pvalue of Z when X = 11.1. So

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

By the Central Limit Theorem

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

$Z = \frac{11.1 - 13.1}{0.82}$

$Z = -2.44$

$Z = -2.44$ has a pvalue of 0.0073.

0.73% probability that their mean length is less than 11.1 inches