Respuesta :
Answer:
1.66% probability that x¯<1.57.
Step-by-step explanation:
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.
In this problem, we have that:
[tex]\mu = 1.59, \sigma = 0.042, n = 20, s = \frac{0.042}{\sqrt{20}} = 0.0094[/tex]
What is the probability that x¯<1.57?
This probability is the pvalue of Z when [tex]X = 1.57[/tex]. So:
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1.57-1.59}{0.0094}[/tex]
[tex]Z = -2.13[/tex]
[tex]Z = -2.13[/tex] has a pvalue of 0.0166.
So there is a 1.66% probability that x¯<1.57.
