Answer:
40.13% probability that 9 randomly selected packages will have an average weight in excess of 16.025 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 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 normally distributed variable X, with mean and standard deviation
, the sample means with size n can be approximated to a normal distribution with mean
and standard deviation
In this problem, we have that:
What is the probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces
This is 1 subtracted by the pvalue of Z when X = 16.025. So
By the Central Limit Theorem
has a pvalue of 0.5987
1 - 0.5987 = 0.4013
40.13% probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces