Respuesta :
Answer:
0.89% probability that a palette of 10 boxes will exceed that limit (150 lbs.)
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 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 \mu and standard deviation, which is also called standard error [tex]s = \frac{\sigma}{\sqrt{n}}[/tex]
In this problem, we have that:
[tex]\mu = 12, \sigma = 4, n = 10, s = \frac{4}{\sqrt{10}} = 1.2649[/tex]
1. What’s the probability that a palette of 10 boxes will exceed that limit (150 lbs.)? a. HINT: Prob (total weight of a sample of 10 boxes >150 lbs.) = Prob (the sample mean weight ( y hat ) > 15 lbs.)
This is 1 subtracted by the pvalue of Z when X = 15. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{15 - 12}{1.2649}[/tex]
[tex]Z = 2.37[/tex]
[tex]Z = 2.37[/tex] has a pvalue of 0.9911
1 - 0.9911 = 0.0089
0.89% probability that a palette of 10 boxes will exceed that limit (150 lbs.)
