Final answer:
The proportion of values between z-scores of –2.00 and +2.00 in a normal distribution is about 95%, not 68%. The 68% corresponds to the proportion of data within one standard deviation from the mean, or between z-scores of –1 and +1, according to the empirical rule.
Step-by-step explanation:
The statement that the proportion located between z = –2.00 and z = 2.00 in a normal distribution is p = 68% is false. According to the empirical rule, sometimes referred to as the 68-95-99.7 rule, approximately 95% of the values lie between z-scores of –2 and +2. Specifically, the region within one standard deviation of the mean (between z-scores of –1 and +1) contains about 68% of the values. However, the region within two standard deviations (between z-scores of –2 and +2) encompasses about 95% of the data in a normal distribution.
This concept can also be illustrated by considering a normal distribution with a mean (μ) and standard deviation (σ). As an example, suppose x has a normal distribution with mean 50 and standard deviation 6. About 68 percent of the x values lie within one standard deviation (between 44 and 56), which corresponds to z-scores of –1 and +1, while about 95% fall within two standard deviations (between 38 and 62), corresponding to z-scores of –2 and +2.