Respuesta :
Using the normal distribution and the central limit theorem, it is found that there is a 0.0028 = 0.28% probability that the manufacturing line will be shut down unnecessarily.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It 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, which is the percentile of X.
- By the Central Limit Theorem, the sampling distribution of sample means of size n have standard deviation given by [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this problem:
- Mean of 0.75 inches, thus [tex]\mu = 0.75[/tex].
- Standard deviation of 0.04 inches, thus [tex]\sigma = 0.04[/tex]
- Sample of 36, thus [tex]n = 36, s = \frac{0.04}{\sqrt{36}} = 0.0067[/tex].
The probability of being shutdown is the probability of a sample mean less than 0.73 inches or more than 0.77 inches. They are the same distance to the mean(0.02), thus, considering the Central Limit Theorem:
[tex]z = \frac{0.02}{s}[/tex]
[tex]z = \frac{0.02}{0.0067}[/tex]
[tex]z = 2.99[/tex]
The probability is P(|z| > 2.99), which is 2 multiplied by the p-value of z = -2.99.
z = -2.99 has a p-value of 0.0014.
2(0.0014) = 0.0028
0.0028 = 0.28% probability that the manufacturing line will be shut down unnecessarily.
A similar problem is given at https://brainly.com/question/25212662
