The mean weight of boxes shipped by a company is 12 lbs., with a standard deviation of 4 lbs. Assume weight of boxes follows Normal Distribution. Boxes are shipped in palettes of 10 boxes. The shipper has a limit of 150 lbs. for such shipments. 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.)

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation , a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error

In this problem, we have that:

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

By the Central Limit Theorem

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.)

b. 153 students do not participate in sports and seventy-five 12th graders were surveyed