The population distribution is normal with a mean of .50 and a standard deviation of .20.

First we calculate the standardized z score corresponding to a deviation of .05 from the mean to be z = (x - m )/s = (.55 - .50)/.20 = .05/.20 = .25. Then we use the z table to find that the probability that a randomly selected score has a z value between zero and .25 is .0987. Next, to find the probability of observing a z score between -.25 and +.25, we double .0987 to get .1974, or about 20%.

If we knew the standard deviation of the sampling distribution of the mean, we could calculate the z score for any observed mean and then use z tables to find the probability of observing a mean within any specific range. Fortunately, from the Central Limit Theorem we know that the standard error of the mean (i.e., the standard deviation of the sampling distribution of the mean) is equal to the standard deviation of the population divided by the square root of the sample size. In our example, the standard error of the mean is .20 / (square root of 100) = .20/10 = .020. Thus, a sample mean that is .02 greater than the population mean has a z score of 1.00.

In our example, we would like to know how likely it is that a sample mean differs from the population mean by .05. This corresponds to a z score of .05/.02 = 2.50. The probability of observing a sample mean between .45 and .55 in our example is the same as the probability of observing a z score between -2.50 and +2.50 on a standardized normal distribution. From the z table we find the probability of observing a z score greater than 2.50 is only .0062; the probability of observing a z score less than -2.50 is also .0062. Thus, the probability of observing a z score between -2.50 and +2.50 is 1.0000 - .0062 - .0062 = .9876.

We conclude that the probability is over 98% that a randomly selected sample of n=100 American adults will have a mean Life Satisfaction score within .05 of the actual population mean.

The standard error of the mean is .20 / (square root of 5) = .20/2.236 = .08944. A sample mean that is .05 greater than the population mean has a z score of .05/.08944 = .56. From the z table we find the probability of observing a z score greater than .56 is .2877. Thus, the probability of observing a z score between -.56 and +.56 is 1.0000 - .2877 - .2877 = .4246.

We conclude that the probability is about 42% that a randomly selected sample of n=5 American adults will have a mean Life Satisfaction score within .05 of the actual population mean, and a 58% chance that the error in estimate is greater than .05.

• Go to the Sampling Distributions applet
• An overview of the applet
• Exercise 1: Accuracy of sample means
• Introduction to the Applet
• Review of z-scores
• Review of Central Limit Theorem
• Answers for Exercise 1 hand calculation problems
• Follow up questions: Questions to assess knowledge
• Download the exercise in Rich Text Format or as an Adobe Acrobat pdf

Questions, comments, difficulties? Please contact Dale Berger.