Answer :
Answer:
22.66% probability that the mean weight will exceed 3.81 ounces
Step-by-step explanation:
To solve this question, we have to understand the normal probability distribution and the central limit theorem.
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.
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 = 3.78, \sigma = 0.14, n = 12, s = \frac{0.14}{\sqrt{12}} = 0.04[/tex]
What is the probability that the mean weight will exceed 3.81 ounces
This probability is 1 subtracted by the pvalue of Z when X = 3.81. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{3.81 - 3.78}{0.04}[/tex]
[tex]Z = 0.75[/tex]
[tex]Z = 0.75[/tex] has a pvalue of 0.7734
1 - 0.7734 = 0.2266
22.66% probability that the mean weight will exceed 3.81 ounces