Question

A synthetic fiber used in manufacturing carpet has a tensile strength that is normally distributed with a mean 75.5 psi and standard deviation 3.5 psi. Find the probability that a random sample of n = 6 fiber specimens will have a sample tensile strength that exceeds 75.75 psi.

Answer 1

Answer:

43.25% probability that a random sample of n = 6 fiber specimens will have a sample tensile strength that exceeds 75.75 psi.

Step-by-step explanation:

To solve this problem, 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 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 $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 75.5, \sigma = 3.5, n = 6, s = \frac{3.5}{\sqrt{6}} = 1.43$

Find the probability that a random sample of n = 6 fiber specimens will have a sample tensile strength that exceeds 75.75 psi.

This is 1 subtracted by the pvalue of Z when X = 75.75. So

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

$Z = \frac{75.75 - 75.5}{1.43}$

$Z = 0.17$

$Z = 0.17$ has a pvalue of 0.5675.

1 - 0.5675 = 0.4325

43.25% probability that a random sample of n = 6 fiber specimens will have a sample tensile strength that exceeds 75.75 psi.