Question

The mean cost of a five pound bag of shrimp is 50 dollars with a variance of 64. If a sample of 43 bags of shrimp is randomly selected, what is the probability that the sample mean would differ from the true mean by greater than 1 dollar? Round your answer to four decimal places.

Answer 1

Answer:

0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar

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 $\mu$ and standard deviation(which is the square root of the variance) $\sigma$, 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 random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 50, \sigma = \sqrt{64} = 8, n = 43, s = \frac{8}{\sqrt{43}} = 1.22$

What is the probability that the sample mean would differ from the true mean by greater than 1 dollar?

Either it differs by 1 dollar or less, or it differs by more than one dollar. The sum of the probabilities of these events is decimal 1.

Probability it differs by 1 dollar or less:

pvalue of Z when X = 50+1 = 51 subtracted by the pvalue of Z when X = 50 - 1 = 49.

X = 51

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

By the Central Limit Theorem

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

$Z = \frac{51-50}{1.22}$

$Z = 0.82$

$Z = 0.82$ has a pvalue of 0.7939

X = 49

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

$Z = \frac{49-50}{1.22}$

$Z = -0.82$

$Z = -0.82$ has a pvalue of 0.2061

0.7939 - 0.2061 = 0.5878

Probability it differs by more than 1 dollar:

p + 0.5878 = 1

p = 0.4122

0.4122 = 41.22% probability that the sample mean would differ from the true mean by greater than 1 dollar