Packages of sugar bags for Sweeter Sugar Inc. have an average weight of 16 ounces and a standard deviation of 0.3 ounces. The weights of the sugar packages are normally distributed. What is the probability that 9 randomly selected packages will have an average weight in excess of 16.025 ounces

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

the answer is (-4,-4)

hope this ! : d