Respuesta :
Answer:
0.62% probability that the sample mean weight of these 100 bags is less than 10.45 ounces.
Step-by-step explanation:
To solve this question, the concepts of the normal probability distribution and the central limit theorem are important.
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]\frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 10.5, \sigma = 0.2, n = 100, s = \frac{0.2}{\sqrt{100}} = 0.02[/tex]
Find the probability that the sample mean weight of these 100 bags is less than 10.45 ounces
This is the pvalue of Z when X = 10.45. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{10.45 - 10.5}{0.02}[/tex]
[tex]Z = -2.5[/tex]
[tex]Z = -2.5[/tex] has a pvalue of 0.0062.
So there is a 0.62% probability that the sample mean weight of these 100 bags is less than 10.45 ounces.
