Answer:
0.9452 = 94.52% probability that the sample mean would be less than 107.81 liters.
Step-by-step explanation:
To solve this question, 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
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.
Central limit theorem:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 105, \sigma = 26, n = 220, s = \frac{26}{\sqrt{220}} = 1.7529](https://tex.z-dn.net/?f=%5Cmu%20%3D%20105%2C%20%5Csigma%20%3D%2026%2C%20n%20%3D%20220%2C%20s%20%3D%20%5Cfrac%7B26%7D%7B%5Csqrt%7B220%7D%7D%20%3D%201.7529)
Probability that the sample mean would be less than 107.81 liters?
pvalue of Z when X = 107.81. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the central limit theorem
![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{107.81 - 105}{1.7529}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B107.81%20-%20105%7D%7B1.7529%7D)
![Z = 1.60](https://tex.z-dn.net/?f=Z%20%3D%201.60)
has a pvalue of 0.9452
0.9452 = 94.52% probability that the sample mean would be less than 107.81 liters.