Final answer:
To find the range of sample mean earnings for 40 shifts at a restaurant with 98% probability, calculate the standard error of the mean and apply a z-score corresponding to the 98% confidence level. The final range is centered around the mean of $248 and extends by the z-score multiplied by the SEM in both directions.
Step-by-step explanation:
The question is asking to determine the range within which the sample mean earnings for a restaurant's waitstaff is expected to fall during 40 shifts, with a 98% probability. Given that the average earnings per shift are $248 with a standard deviation of $39, we need to apply the Central Limit Theorem to calculate the interval for the sample mean. The Central Limit Theorem explains that the distribution of the sample means will be approximately normally distributed, even if the population distribution is not normal, provided the sample size is large enough (usually n>30 is considered sufficient).
First, we calculate the standard error of the mean (SEM), which is the standard deviation of the sample mean distribution:
SEM = σ/ √ n = $39 / √ 40.
Substituting the given values gives us the SEM, and then we can find the z-score corresponding to a 98% confidence level. The z-score for 98% confidence is approximately 2.33. Using the z-score, we can calculate the range:
Range = ± z * SEM. With the z-score and the SEM, we have the lower and upper bounds of the interval centered around the mean of $248.
The range in which they can expect the sample mean to fall with a 98% probability is from $248 minus the product of 2.33 and SEM, to $248 plus the same product. This completes the calculation for the 98% confidence interval. It is important for students to understand the application of the Central Limit Theorem when dealing with sample means and how the standard error and z-scores play a vital role in constructing confidence intervals.