Respuesta :
Answer:
a) 79.67% probability that a randomly selected bottle has at least 355 ml
b) 99.29% probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml.
c) 88.88% probability that the total amount of beer in the 6-pack is less than 2131.5 ml
Step-by-step explanation:
To solve this question, we need to understand the normal probability distribution and the central limit theorem.
Normal probability distribution
When the distribution is normal, we use 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 normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
In this question:
[tex]\mu = 355.5, \sigma = 0.6[/tex]
a. What is the probability that a randomly selected bottle has at least 355 ml?
This is 1 subtracted by the pvalue of Z when X = 355. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{355 - 355.5}{0.6}[/tex]
[tex]Z = -0.83[/tex]
[tex]Z = -0.83[/tex] has a pvalue of 0.2033
1 - 0.2033 = 0.7967
79.67% probability that a randomly selected bottle has at least 355 ml.
b. What is the probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml?
Now [tex]n = 6, s = \frac{0.5}{\sqrt{6}} = 0.2041[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{355 - 355.5}{0.2041}[/tex]
[tex]Z = -2.45[/tex]
[tex]Z = -2.45[/tex] has a pvalue of 0.0071
1 - 0.0071 = 0.9929
99.29% probability that the bottles of a randomly selected 6-pack of beer have an average of at least 355 ml.
c. What is the probability that the total amount of beer in the 6-pack is less than 2131.5 ml?
2131.5/6 = 355.25
This is the pvalue of Z when X = 355.25.
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{355.25 - 355.5}{0.2041}[/tex]
[tex]Z = -1.22[/tex]
[tex]Z = -1.22[/tex] has a pvalue of 0.1112
1 - 0.1112 = 0.8888
88.88% probability that the total amount of beer in the 6-pack is less than 2131.5 ml
