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
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation ![\frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
![\mu = 1.59, \sigma = 0.042, n = 20, s = \frac{0.042}{\sqrt{20}} = 0.0094](https://tex.z-dn.net/?f=%5Cmu%20%3D%201.59%2C%20%5Csigma%20%3D%200.042%2C%20n%20%3D%2020%2C%20s%20%3D%20%5Cfrac%7B0.042%7D%7B%5Csqrt%7B20%7D%7D%20%3D%200.0094)
What is the probability that x¯<1.57?
This probability is the pvalue of Z when
. So:
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{1.57-1.59}{0.0094}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1.57-1.59%7D%7B0.0094%7D)
![Z = -2.13](https://tex.z-dn.net/?f=Z%20%3D%20-2.13)
has a pvalue of 0.0166.
So there is a 1.66% probability that x¯<1.57.