The tip percentage at a restaurant has a mean value of 18% and a standard deviation of 6%.

a. What is the approximate probability that the sample mean tip percentage for a random sample of 40 bills is between 16% and 19%.
b. If the sample size had been 15 rather than 40, could the probability requested in part (a) be calculated from the given information?

We will use the central limit theorem which can be applied to a random sample from any distribution as long as the mean and the variance are both finite and the sample size is large (the sample size n should be greater than 30). Here we have that the tip percentage at the restaurant has a mean value of 18% and a standard deviation of 6%, then because of the central limit theorem, we know that the sample mean tip percentage is approximately normally distributed with mean 18% and standard deviation [tex]6/\sqrt{40}%=0.9487%[/tex].

a. The z-score related to 16% is given by (16-18)/0.9487 = -2.1081 and the z-score related to 19% is given by (19-18)/0.9487 = 1.0541. We are looking for [tex]P(16 < \bar{X} < 19) = P(-2.1081 < Z < 1.0541) = P(Z < 1.0541)- P(Z < -2.1081) = 0.8540 - 0.0175 = 0.8366[/tex]

b. If the sample size had been 15 rather than 40, then, the probability requested in part (a) could not be calculated from the given information, this because the central limit theorem only applies when the sample size is large, for example n > 30.