2. By definition, jumbo shrimp are those that require between 10 and 15 shrimp to make a pound. Suppose that the number of jumbo shrimp in a 1-pound bag averages μ = 12.5 with a standard deviation of σ = 1.5, and forms a normal distribution. What is the probability of randomly picking a sample of n =25, 1-pound bags that average more than M = 13 shrimp per bag?

0.0475 = 4.75% probability of randomly picking a sample of n =25, 1-pound bags that average more than M = 13 shrimp per bag.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal Probability Distribution:

Problems of normal distributions 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 normally distributed random variable X, with mean and standard deviation , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean and standard deviation .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

Single bag:

Sample of 25 bags:

What is the probability of randomly picking a sample of n =25, 1-pound bags that average more than M = 13 shrimp per bag?

This is 1 subtracted by the pvalue of Z when X = 13. So

By the Central Limit Theorem

has a pvalue of 0.9525

1 - 0.9525 = 0.0475

0.0475 = 4.75% probability of randomly picking a sample of n =25, 1-pound bags that average more than M = 13 shrimp per bag.

-2 is the right