Respuesta :
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 [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 = 10, \sigma = 0.1, n = 625, s = \frac{0.1}{\sqrt{625}} = 0.004[/tex]
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
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{9.995 - 10}{0.004}[/tex]
[tex]Z = -1.25[/tex]
[tex]Z = -1.25[/tex] 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.
