Final answer:
The probability that the sample mean of the cost of shrimp differs from the true mean by less than $1.4 is found using the standard error and z-scores. A z-score is calculated for the difference of $1.4 and the probability is determined by the area under the standard normal distribution curve between z-scores -1.54 and 1.54.
Step-by-step explanation:
To find the probability that the sample mean will differ from the true mean by less than $1.4 in a sample of 43 bags of shrimp, we need to use the sample distribution of the mean. The true mean cost of a five-pound bag of shrimp is given as 47 dollars, with a variance of 36 dollars2. This means the standard deviation is the square root of the variance, which is 6 dollars.
When we take a sample, the standard deviation of the sample mean (also known as the standard error) is the standard deviation divided by the square root of the sample size. Therefore, the standard error (SE) for a sample of 43 bags is:
SE = σ / √n = 6 / √43 ≈ 0.91 dollars
We are looking for the probability that the sample mean is within 1.4 dollars of the population mean. To find this, we calculate the z-score for the difference of 1.4 dollars:
Z = (X - μ) / SE = 1.4 / 0.91 ≈ 1.54
We will then use the standard normal distribution to find the probability that the z-score is less than 1.54 and greater than -1.54 (representing a difference less than 1.4 dollars in both directions from the mean). This can be found using a standard normal distribution table or a calculator with cumulative distribution function capabilities.
The final answer is the combined area under the standard normal curve between -1.54 and 1.54. This area represents the probability that the sample mean is within 1.4 dollars of the true mean.