Consider the performance analysis problem where you are interested in the service time (y) of database transactions. Let us assume that the mean for this population (i.e. mean service time) is mu=100ms with a standard deviation of sigma=20ms. Considering the sampling distribution of ybar, answer the following: a) For samples of size n=400, what percentage of samples would you expect to result in sample means between 98ms and 102ms. Soln: Since the sample size is large (i.e. n>=30) we know that mu(ybar)=mu=100 and sigma(ybar)=sigma/sqrt(n)=20/sqrt(400)=1. We also know that the ybars are normally distributed. Hence, using the transformations rule z=(x-mu(ybar))/sigma(ybar), z1=(98-100)/1=-2 and z2=(102-100)/1=2. From the empirical rule we know that the desired proportion is 95.45%. b) For samples of size n=400, what percentage of samples would you expect to result in sample means greater than 103ms. Soln: Again, the sample size is large (i.e. n>=30) and so, as in a), mu(ybar)=mu=100 and sigma(ybar)=sigma/sqrt(n)=20/sqrt(400)=1. We also know that the ybars are normally distributed. Hence z=(x-mu(ybar))/sigma(ybar)=(103-100)/1=3. From the empirical rule we know that the desired proportion is (50-99.73/2)=0.135%. c) In this case, let the sample size n=26. What percentage of samples would you now expect to result in sample means greater than 103ms. Make any appropriate assumptions. Soln: The sample size is small (i.e. n<30) and so we know that mu(ybar)=mu=100 and sigma(ybar)=sigma/sqrt(n-1)=20/sqrt(25)=4. We also know that the ybars will be Student t distributed with n-1=25 degrees of freedom if we assume that the service times (i.e. y) are normally distributed. Hence, with this assumption, t=(x-mu(ybar))/sigma(ybar)=(103-100)/4=0.75. From the t table, traversing the row for df=25, 0.75 is between 0.68 and 1.32, hence the desired proportion is between 10% and 25%.