The mean cost of a five pound bag of shrimp is 50 dollars with a standard deviation of 8 dollars. If a sample of 56 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 0.9 dollars

59.90% probability that the sample mean would differ from the true mean by less than 0.9 dollars

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:

If a sample of 56 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by less than 0.9 dollars

This is the pvalue of Z when X = 50 + 0.9 = 50.9 subtracted by the pvalue of Z when X = 50 - 0.9 = 49.1. So

X = 50.9

By the Central Limit Theorem

has a pvalue of 0.7995

X = 49.1

has a pvalue of 0.2005

0.7995 - 0.2005 = 0.5990

59.90% probability that the sample mean would differ from the true mean by less than 0.9 dollars

answer: he is 272195

step-by-step explanation: all you have to do is multiple 3535 and 77 and it equals 272195