Answer :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], a large sample size can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 75.5, \sigma = 3.5, n = 6, s = \frac{3.5}{\sqrt{6}} = 1.43[/tex]
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
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{75.75 - 75.5}{1.43}[/tex]
[tex]Z = 0.17[/tex]
[tex]Z = 0.17[/tex] 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.
43.25% of fiber specimens will have a sample tensile strength that exceeds 75.75 psi.
What is z score?
Z score is used to determine by how many standard deviations the raw score is above or below the mean.
z = (raw score - mean) / (standard deviation ÷ √sample)
mean 75.5 psi and standard deviation 3.5 psi. n = 6.
For > 75.75:
z = (75.75 - 75.5) / (3.5 ÷√6) = 0.17
P(z > 0.17) = 1 - P(z < 0.17) = 1 - 0.5675 = 43.25%
43.25% of fiber specimens will have a sample tensile strength that exceeds 75.75 psi.
Find out more on z score at: https://brainly.com/question/25638875