Respuesta :
Answer:
C. No, the probability of this outcome at 0.080, would be considered usual, so there is no problem
Step-by-step explanation:
To solve this problem, it is important to know the Central Limit Theorem and the Normal probability distribution.
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]\frac{\sigma}{\sqrt{n}}[/tex]
Normal Probability Distribution
Problems of normally distributed samples can be 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.
When a probability is unusual?
A probability is said to be unusual if it is higher than 0.95 or lower than 0.05.
In this problem, we have that:
[tex]\mu = 20, \sigma = 0.11, n = 40, s = \frac{0.11}{\sqrt{15}} = 0.0284[/tex]
Should the machine be reset?
What is the probability of a mean of 20.04 or higher?
This is 1 subtracted by the pvalue of Z when X = 20.04. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{20.04 - 20}{0.0284}[/tex]
[tex]Z = 1.41[/tex]
[tex]Z = 1.41[/tex] has a pvalue of 0.92.
So the probability of this outcome is 1-0.92 = 0.08 = 8%.
8% is still an usual probability.
So the correct answer is:
C. No, the probability of this outcome at 0.080, would be considered usual, so there is no problem
