Answer:
10.57% probability that the mean contents of the 625 sample cans is less than 9.995 ounces.
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
.
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 = 10, \sigma = 0.1, n = 625, s = \frac{0.1}{\sqrt{625}} = 0.004](https://tex.z-dn.net/?f=%5Cmu%20%3D%2010%2C%20%5Csigma%20%3D%200.1%2C%20n%20%3D%20625%2C%20s%20%3D%20%5Cfrac%7B0.1%7D%7B%5Csqrt%7B625%7D%7D%20%3D%200.004)
What is the probability that the mean contents of the 625 sample cans is less than 9.995 ounces?
This is the pvalue of Z when X = 9.995. 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{9.995 - 10}{0.004}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B9.995%20-%2010%7D%7B0.004%7D)
![Z = -1.25](https://tex.z-dn.net/?f=Z%20%3D%20-1.25)
has a pvalue of 0.1057
So there is a 10.57% probability that the mean contents of the 625 sample cans is less than 9.995 ounces.