The cost of 5 gallons of ice cream has a standard deviation of 8 dollars with a mean of 29 dollars during the summer. What is the probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected?

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

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 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 , the sample means with size n of at least 30 can be approximated to a normal distribution with mean and standard deviation

In this problem, we have that:

What is the probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected?

This is the pvalue of Z when X = 29 + 1.9 = 30.9 subtracted by the pvalue of Z when X = 29 - 1.9 = 27.1. So

X = 30.9

By the Central Limit Theorem

has a pvalue of 0.9887

X = 27.1

has a pvalue of 0.0113

0.9887 - 0.0113 = 0.9774

97.74% probability that the sample mean would differ from the true mean by less than 1.9 dollars if a sample of 92 5-gallon pails is randomly selected

i dont know sorry

step-by-step explanation: