Mathematics, 24.03.2020 20:19 jaylanwest12

The mean cost of a five pound bag of shrimp is 40 dollars with a standard deviation of 8 dollars. If a sample of 49 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by greater than 2.6 dollars? Round your answer to four decimal places.

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Answer from: CuteDoggo1828

0.0229 = 2.29% probability that the sample mean would differ from the true mean by greater than 2.6 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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

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

In this problem, we have that:

$\mu = 40, \sigma = 8, n = 49, s = \frac{8}{\sqrt{49}} = 1.14$

What is the probability that the sample mean would differ from the true mean by greater than 2.6 dollars?

Eiter it differs by 2.6 dollars or less, or it differs by more than 2.6 dollars. The sum of the probabilities of these events is decimal 1. So

Probability it differs by 2.6 dollars or less.

pvalue of Z when X = 40+2.6 = 42.6 subtracted by the pvalue of Z when X = 40-2.6 = 37.4.

X = 42.6

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{42.6 - 40}{1.14}$

Z = 2.275

Z = 2.275 has a pvalue of 0.98855

X = 37.4

$Z = \frac{X - \mu}{s}$

$Z = \frac{37.4 - 40}{1.14}$

Z = -2.275

Z = -2.275 has a pvalue of 0.01145

0.98855 - 0.01145 = 0.9771

0.9771 probability it differs by 2.6 dollars or less.

Probability it differs by greater than 2.6 dollars.

p + 0.9771 = 1

p = 0.0229

0.0229 = 2.29% probability that the sample mean would differ from the true mean by greater than 2.6 dollars

Answer from: Quest

C. i possibility the answer to this

Answer from: Quest

step-by-step explanation:

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