A restaurant's receipts show that the cost of customers' dinners has a skewed distribution with a mean of $54 and a standard deviation of $18. Explain, in the context of the question, why you cannot draw and label a sampling model for the mean (average) cost of customers' dinners if we took a random sample of 10 people ABI E PANormal Population Assumption: as stated in the question, the distribution of restaurant receipts is skewed, so this assumption is not met. Is the sample size assumption met? Check the large enough sample condition. AY B I E PA What are the mean and standard deviation for the mean cost of dinner, for the sample of 100 customer receipts? mean = $ standard deviation = $Using your answers from the previous and the 68-95-99.7 rule, sketch a normal curve showing the average dinner cost for a sample of 100 customer receipts. Fill in the ranges below. 68% of the data will fall between $ and $ 95% of the data will fall between $ and $ 99.7% of the data will fall between $ and $ If we were to take a random sample of 125 receipts, would the interval containing the middle 68% of values be wider or narrower than the one found in part (e)? [QUESTION 12]? Explain! 1 ABIDE