Respuesta :
Answer:
a) 0.27% probability that the mean is less than 1995 milliliters.
b) 2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.
Step-by-step explanation:
To solve this problem, it is important 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 [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.
Central Limit theorem
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]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 2000, \sigma = 18[/tex]
A) If the manufacturer samples 100 bottles, what is the probability that the mean is less than 1995 milliliters?
So [tex]n = 100, s = \frac{18}{\sqrt{100}} = 1.8[/tex]
This probability is the pvalue of Z when X = 1995. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{1995 - 2000}{1.8}[/tex]
[tex]Z = -2.78[/tex]
[tex]Z = -2.78[/tex] has a pvalue of 0.0027
So 0.27% probability that the mean is less than 1995 milliliters.
B) What mean overfill or more will occur only 10% of the time for the sample of 100 bottles?
This is the value of X when Z has a pvalue of 1-0.1 = 0.9.
So it is X when [tex]Z = 1.28[/tex]
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]1.28 = \frac{X - 2000}{1.8}[/tex]
[tex]X - 2000 = 1.8*1.28[/tex]
[tex]X = 2002.3[/tex]
2002.3 milliliters or more will occur only 10% of the time for the sample of 100 bottles.
