A café wants to study how customers tip differently based on what they order. The distribution of tips given by customers who buy only a cup of coffee is bimodal with a mean of $0.29 and a standard deviation of $0.116. The distribution of tips given by customers who buy only a salad is approximately Normally distributed with a mean of $2.89 and a standard deviation of $1.18. If a random sample of 35 tips from customers who buy only a cup of coffee is selected and a random sample of 20 customers who buy only a salad is selected, what is the probability of a sample mean being at least $2.50 more for customers who buy only a salad than for those who buy only a cup of coffee?
Let the difference be the salad sample mean minus the coffee sample mean. Its mean is $2.89 − $0.29 = $2.60, and its standard error is √(1.18²/20 + 0.116²/35) ≈ $0.265. Thus, a difference of $2.50 is about (2.50 − 2.60)/0.265 ≈ −0.38 standard errors below the mean. The probability of a difference at least $2.50 is therefore about 0.6473, so the answer is C, 0.6473.
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